I believe that the answer is no (evidenced by the 80+ year "open" status of the conjecture), but I am self-taught in formal logic and decidability theory, so I would like to present my argument for review.
Presburger arithmetic, the first-order theory $Th(\mathbb N, +)$ of natural numbers $\mathbb N = \{0,1,2,\ldots\}$ with addition, is known to be a decidable theory. There is an algorithm that will take any sentence like $\forall x \exists y (x = y + y \texttt{ or } x = y + y + 1)$ and compute whether it or its negation can be deduced from the axioms (though the runtime of the algorithm is $\mathcal O(2^{2^{cn}})$ for a constant $c > 0$, with $n = $ length of the sentence). And the (shortened) Collatz function $T$ is definable in Presburger, by
$$T(x) = y \iff (x = y + y \texttt{ or } x + x + x + 1 = y + y)$$
If it were then possible to define arbitrary iterates of $T$, ie if we had the function $T(n,x) = T^n(x)$, then perhaps it would be possible to use the Presburger decision algorithm to settle the Collatz Conjecture. As it turns out, $T(n,x)$ is not definable in Presburger (argued below), but perhaps there is a decidable expansion of Presburger that is strong enough to state the Collatz Conjecture.
So suppose we augment Presburger with the new function $T(n,x)$, defined by two axioms: $T(0,x) = x$ and $\forall n (T(n+1,x) = T(T(n,x))$. Note that we are using the Presburger function $T(x)$ defined above. Note also that we are not asking for the ability to iterate any Presburger-definable function arbitrarily many times (which would trivially lead to a definition of multiplication and therefore an undecidable theory at least as strong as Peano), but we are asking only for the iterates of the map $T$.
Note: the following relations are definable in Presburger.
- $x \leq y \iff \exists z(x+z = y)$. This takes advantage of the fact that all variables are natural numbers (and thence nonnegative).
- $x \texttt{ is even} \iff \exists y(x = y + y)$. This leads immediately to the relation $x = y \bmod 2$.
- $x = \max(k : \varphi(k)) \iff \big(\varphi(x) \texttt{ and } \forall y > x\ (\texttt{not } \varphi(y))\big)$ for a Presburger-definable predicate $\varphi$. Similarly, $\min(k : \varphi(k))$ is definable.
The Collatz map $T$ has the property that the parity vector $p_n(x) = (T^i(x) \bmod 2)_{i=0}^{n-1}$ of length $n$ is unique to $x \bmod 2^n$. It is therefore possible to define the predicate
$$M(x,y,n) \iff \forall i < n\ \big(T(i,x) = T(i,y) \bmod 2\big)$$
which from the above property is equivalent to $x = y \bmod 2^n$. From this it becomes possible to identify the powers of 2, as well as all multiples of powers of $2$. The power-of-two function may be defined by
$$2 \wedge n = \min(x : x > 0 \texttt{ and } x = 0 \bmod 2^n)$$
Note: This justifies that $Th(\mathbb N, +, T(\cdot,\cdot))$ is a strictly stronger theory than Presburger, because Presburger can only identify semilinear sets (finite unions of arithmetic progressions, plus or minus a finite set), and the powers of two are not a semilinear set.
Also, I do not know if the theory described so far is strong enough to actually prove that the function $2 \wedge n$ given above satisfies the axioms $2 \wedge 0 = 1$ and $2 \wedge (n+1) = (2\wedge n) + (2 \wedge n)$. But it is surely true, as provable in a stronger theory (say, ZFC). So, if "Presburger + Collatz" is not strong enough to prove this, then automatically the theory is undedicable(?). Supposing to the contrary all the true first order statements about $2\wedge n$ are provable here, we proceed.
This also leads to the 2-adic valuation function $V_2(x) = 2 \wedge \max(n : x = 0 \bmod 2^n)$, which takes $x$ to the highest power of $2$ dividing it. All this shows that the "Presburger + Collatz" theory we have described so far is at least as strong as $Th(\mathbb N, +, V_2)$ - a theory known to be decidable (Buchi-Bruyere Theorem). There is yet hope! But the hope is dashed by the following observation.
The Collatz map $T$ also satisfies the identity $T^n(2^n - 1) = 3^n - 1$. Care must be taken to avoid the use of subtraction, but this means it is possible to define the power-of-three function as well.
$$x = 3 \wedge n \iff \exists y(y + 1 = 2 \wedge n \texttt{ and } x = T(n, y) + 1)$$
Therefore Pr + Cz is at least as strong as $Th(\mathbb N, +, V_2, P_3)$, shown to be undedicable by A. Bes here.
Edit: The Collatz map $T$ also satisfies the identity $T^n(2^nk - 1) = 3^nk - 1$ for any $n > 1$, $k \in \mathbb Z$, so with a little additional work it becomes possible to identify not only the powers of $3$, but also the multiples of all powers of $3$.
$$x \texttt{ is a multiple of } 3^n \iff (x = 0) \texttt{ or } \exists y\big(y + 1 = 0 \bmod 2^n \texttt{ and } x = T(n,y) + 1\big)$$
From this we recover the function $V_3(x) = 3 \wedge \max(n : x \texttt{ is a multiple of } 3^n)$, therefore $(\mathbb N, +, T)$ is at least as strong as $(\mathbb N, +, V_2, V_3)$, known to be undecidable from even earlier (shown by Villemaire here).
