We cannot deduce that $f$ is uniformly continuous
Preparations
Choose a conditionally convergent series $\sum_{k=0}^{\infty} a_k$, e.g.
$$\sum_{k=0}^{\infty} a_k = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \frac{1}{5} - \frac{1}{6} + \frac{1}{7} - \frac{1}{8} + \ldots$$
Given $n \in \mathbb{Z}$, let $b_n$ denote the sum of all $a_k$ such that the ($2$-adic) binary expansion of $n$ has a $1$ on position $k$. In symbols:
$$b_n = \sum_{k=0}^{\infty} c_k \cdot a_k, \qquad \text{where in } \mathbb{Z}_2:\qquad n = \sum_{k=0}^{\infty} c_k \cdot 2^k, c_k \in \{ 0, 1 \}.$$
Note that for negative $n$ ultimately $c_k = 1$, so the series defining $b_n$ converges to a real number.
Claim. $b_{2^k+n} - b_n \xrightarrow{k \to \infty} 0$ uniformly with respect to $n$.
Proof. The binary expansions of $n$ and $n+2^k$ differ in the following way, where $k \leqslant m \leqslant \infty$:
$$\begin{align*}
n &= *** \ \overset{\substack{m \\ \downarrow \\ \phantom{0}}}{0}11 \ldots \overset{\substack{k \\ \downarrow \\ \phantom{0}}}{1} \ *\mathrel{*}\overset{\substack{0 \\ \downarrow \\ \phantom{0}}}{*} \\
2^k+n &= *** \ \underset{\substack{\phantom{0} \\ \uparrow \\ m}}{1}00 \ldots \underset{\substack{\phantom{0} \\ \uparrow \\ k}}{0} \ *\mathrel{*}\underset{\substack{\phantom{0} \\ \uparrow \\ 0}}{*}
\end{align*}$$
It follows that if $m < \infty$,
$$\left| b_{2^k+n} - b_n \right| \leqslant |a_m| + \left| \sum_{i=k}^{m-1} a_i \right|$$
and the right hand side approaches zero because of the Cauchy condition.
If $m = \infty$, the reasoning is similar. $\square$
Counterexample
Define $f(x) = b_n \cdot \sin^2( \pi x )$ on each interval $[n, n+1]$ with $n \in \mathbb{Z}$.
Clearly $f \equiv 0$ on the integers where the piecewise definitions meet, so $f$ is continuous.
For any $k \in \mathbb{N}$, $n \in \mathbb{Z}$ and $x \in [n, n+1]$ we have
$$|f(x+2^k) - f(x)| = |b_{n+2^k} \sin^2 \left( \pi \big( x + 2^k \big) \right) - b_{n} \sin^2( \pi x )| = |b_{n+2^k} - b_n| \cdot \sin^2( \pi x ).$$
It now follows from the Claim that $f$ satisfies the assumption: given $\varepsilon > 0$, take $\delta = 2^k$ where $k$ is so large that $|b_{n+2^k} - b_n| < \varepsilon$ for all $n \in \mathbb{Z}$.
$f$ is not uniformly continuous because $(b_n)$ is unbounded.
Thus the implication does not hold.
