The answer is no, and we give some counter-examples below: two in $2$ and one in $3$ and more dimensions. References for the two-dimensional examples are this paper and this paper, and the three-dimensional example also appears in a slightly different form in these old lecture notes of M. Wolf (Example 8.1). However, in my opinion, these examples are rather difficult to find if you're not already familiar with these references---hence why I decided to create this post.
Interestingly, the second paper linked above shows that---while not all channels can be diagonalized---every channel can be approximated by diagonalizable channels arbitrarily well.
1. The Qubit Case
1a. Example 1
First we will construct some examples based on the Pauli transfer matrix, which is just the representation matrix with respect to the orthonormal basis $\{\frac1{\sqrt2}{\bf1},\frac1{\sqrt2}\sigma_x,\frac1{\sqrt2}\sigma_y,\frac1{\sqrt2}\sigma_z \}$. The idea here is that the Pauli transfer matrix of a unital (i.e. identity-preserving) qubit channel is always of the form
$$
\begin{pmatrix}
1&0\\0&\Lambda
\end{pmatrix}\tag{1}
$$
for some $\Lambda\in\mathbb R^{3\times 3}$. Interestingly, the Fujiwara-Algoet conditions (arXiv)---which in the unital case are necessary and sufficient conditions on the singular values of $\Lambda$ such that (1) gives rise to a quantum channel---show that for every $\Lambda\in\mathbb R^{3\times 3}$ there exists $c>0$ such that the block-diagonal matrix $1\oplus c\Lambda$ corresponds to a unital channel. This gives rise to a simple machinery to construct non-diagonalizable channels:
Starting from your favorite non-diagonalizable real $3\times 3$-matrix $\Lambda$, the channel with Pauli transfer matrix $1\oplus c\Lambda$ (for all $c$ "small enough") is not diagonalizable.
To show this construction in action, one of the simplest examples then is the channel with Pauli transfer matrix
$$
\begin{pmatrix}
1&0&0&0\\0&0&1/2&0\\0&0&0&1/2\\0&0&0&0
\end{pmatrix}
$$
is obviously not diagonalizable. Complete positivity of the corresponding (trace-preserving) linear map
$$
\Phi\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}:=
\begin{pmatrix}
\frac{a_{11}+a_{22}}{2} & \frac{1}{4} i (-a_{11}+a_{12}-a_{21}+a_{22}) \\
\frac{1}{4} i (a_{11}+a_{12}-a_{21}-a_{22}) & \frac{a_{11}+a_{22}}{2} \end{pmatrix}
$$
is due to the fact that its Choi matrix
$$
\frac14\begin{pmatrix}
2 & -i & 0 & i \\
i & 2 & i & 0 \\
0 & -i & 2 & i \\
-i & 0 & -i & 2
\end{pmatrix}
$$
has eigenvalues $1$, $1/2$ (2-fold), and $0$.
Geometrically, the effect of this three-dimensional Jordan block is that
- first, $\Phi$ maps the Bloch ball to a smaller circle in the $y-z-$plane,
- a second application of $\Phi$ yields (part of) the $z$-axis,
- and applying $\Phi$ a third (and more) times always yields the Bloch vector zero.
This is readily verified by the action of $\Phi$ on any Bloch vector:
$
\Phi:(x,y,z)^\top\mapsto (y/2,z/2,0)^\top
$.
In contrast, if the eigenvalue $0$ had geometric multiplicity three (while the identity matrix stays the fixed point of $\Phi$) it would not take three, but only one application of $\Phi$ to map the whole Bloch ball to the origin.
1b. Example 2 (minimal Kraus rank)
Next, two examples where we focus more on the Kraus rank, and we try to make it as small as possible.
The first one is again a unital channel, now of Kraus rank 2 which has appeared in the literature before; its Pauli transfer matrix reads
$$
\begin{pmatrix}
1&0&0&0\\0&0&0&-1\\0&0&0&0\\0&0&0&0
\end{pmatrix}\,,
$$
and it can be written as dephasing followed by the unitary channel generated by
$$
\frac1{\sqrt2}\begin{pmatrix}-1&1\\1&1\end{pmatrix}\,.
$$
To show that diagonalizability can also fail for non-unital channels consider $\Phi:\mathbb C^{2\times 2}\to\mathbb C^{2\times 2}$ defined via
$$
\Phi\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}:=
\begin{pmatrix}
a_{11}+\frac23a_{22}&\frac13a_{12}+\frac13a_{22}\\
\frac13a_{21}+\frac13a_{22}&\frac13a_{22}
\end{pmatrix}\,.
$$
Again, $\Phi$ is obviously trace preserving and $\Phi$ is completely positive because its Choi matrix
$$
C(\Phi)=\begin{pmatrix}
1&0&0&1/3\\
0&0&0&0\\
0&0&2/3&1/3\\
1/3&0&1/3&1/3
\end{pmatrix}
$$
has three positive ($\approx 1.177,0.782,0.040$) and one zero eigenvalue.
Even better, the representation matrix of $\Phi$ is already in upper triangular form so we can read off that it has a simple eigenvalue $1$ and a three-fold eigenvalue $1/3$:
$$
\hat\Phi=\begin{pmatrix}
1&0&0&2/3\\
0&1/3&0&1/3\\
0&0&1/3&1/3\\
0&0&0&1/3
\end{pmatrix}
$$
This triangular form, however, already lets one suspect that the geometric multiplicity of the eigenvalue $1/3$ is less than $3$. And indeed, its Jordan normal form reads $\hat\Phi=SJS^{-1}$ with
$$
J=\begin{pmatrix}
1 & 0 & 0 & 0 \\
0 & \frac{1}{3} & 0 & 0 \\
0 & 0 & \frac{1}{3} & 1 \\
0 & 0 & 0 & \frac{1}{3}
\end{pmatrix}\quad\text{ and }\quad S=
\begin{pmatrix}
1 & 0 & 0 & -1 \\
0 & 0 & \frac{1}{3} & 0 \\
0 & 1 & \frac{1}{3} & 0 \\
0 & 0 & 0 & 1
\end{pmatrix}\,.
$$
Note that this can be used to construct a family of non-diagonalizable channels: simply replace the last column of $\hat\Phi$ by $(2/3,b,b^*,1/3)^T$ for any $b\in\mathbb C\setminus\{0\}$. Then $\hat\Phi$ is not diagonalizable, and $\Phi$ is a channel (completely positive) if and only if $|b|\leq\frac{2\sqrt3}9$.
In this limiting case we obtain the in some sense simplest possible example of a non-diagonalizable channel:
$$
\Phi\begin{pmatrix}a_{11}&a_{12}\\a_{21}&a_{22}\end{pmatrix}:=
\begin{pmatrix}
a_{11}+\frac{2 a_{22}}{3} & \frac{1}{9} \left(3 a_{12}+2 \sqrt{3} a_{22}\right) \\
\frac{1}{9} \left(3 a_{21}+2 \sqrt{3} a_{22}\right) & \frac{a_{22}}{3}
\end{pmatrix}
$$
is a channel with Kraus rank 2 (a choice of Kraus operators is
$$
K_1=\sqrt{\frac{3 +\sqrt3}{18}}\begin{pmatrix} \sqrt3&\sqrt3-1\\0&1 \end{pmatrix}\quad\text{ and }\quad K_2=\sqrt{\frac{3-\sqrt3}{18}}\begin{pmatrix} -\sqrt3&1+\sqrt3\\0&1 \end{pmatrix}
$$
as is readily verified)
but $\hat\Phi$ is not diagonalizable per our previous considerations.
2. Higher Dimensions: Classical Channels
Consider the map $\Phi:\mathbb C^{3\times 3}\to\mathbb C^{3\times 3}$ defined via
$$
\Phi\begin{pmatrix}
a_{11}&a_{12}&a_{13}\\
a_{21}&a_{22}&a_{23}\\
a_{31}&a_{32}&a_{33}
\end{pmatrix}:=
\begin{pmatrix}
a_{11}+\frac23a_{22}+\frac13a_{33}&0&0\\
0&\frac13a_{22}+\frac13a_{33}&0\\
0&0&\frac13a_{33}
\end{pmatrix}
$$
so $\Phi$ is even a classical channel, hence the Choi matrix of $\Phi$ is already diagonal.
Using equation (1) it is easy to see that
$$
\hat\Phi=\begin{pmatrix}
1&0&0&0&\frac23&0&0&0&\frac13 \\
0&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0 \\
0&0&0&0&\frac13&0&0&0&\frac13 \\
0&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&0 \\
0&0&0&0&0&0&0&0&\frac13
\end{pmatrix}\simeq
\frac13\begin{pmatrix}
3&2&1\\
0&1&1\\
0&0&1
\end{pmatrix}\oplus 0_6\ .
$$
Now the non-trivial part of $\hat\Phi$ (which we will denote by $D_\Phi$ for "diagonal action") is not diagonalizable; indeed the Jordan normal form of $D_\Phi$ reads $SJS^{-1}$ where
$$
J=\begin{pmatrix}
3&0&0\\
0&1&1\\
0&0&1
\end{pmatrix}\quad\text{ and }\quad S=\begin{pmatrix}
1&-1&-1\\
0&1&0\\
0&0&1
\end{pmatrix}\quad\text{ so }\quad
S^{-1}=\begin{pmatrix}
1&1&1\\
0&1&0\\
0&0&1
\end{pmatrix}\,.
$$
For more examples of non-diagonalizable stochastic matrices refer, e.g., to this math.SE answer.
Moreover, the structure of this counterexample generalizes to any dimension $n\geq 3$.
Addendum: if trace-preservation is dropped, finding a counterexample is a lot easier. Indeed, the—obviously non-diagonalizable—matrix
$$
\begin{pmatrix}0&0&0&1\\0&0&0&0\\0&0&0&0\\0&0&0&0\end{pmatrix}
$$
is the superoperator corresponding to the map
$$
\Phi(X):=\begin{pmatrix}x_{22}&0\\0&0\end{pmatrix}\,,
$$
and $\Phi$ is completely positive as $\{|0\rangle\langle 1|\}$ is a set of Kraus operators of $\Phi$. However, $\Phi$ is obviously not trace-preserving.
