Yes. Surprisingly, $\operatorname{Hom}(\prod_I \mathbb{Z}, \mathbb{Z})\cong \bigoplus_I \mathbb{Z}$, so you can dualize to the corresponding statement for direct sums, and then dualize back.

The key statement is that a homomorphism $\prod_I \mathbb{Z} \to \mathbb{Z} $ necessarily has "finite support". This can be reduced to the case $I=\mathbb{N}$ with a retract argument, and then proceeds by contradiction: one may assume $f(e_i)\neq 0$ for all the "standard basis vectors" $e_i$, and then looks at elements of the form $x_n = \sum_{i=0}^n a_i e_i$ with $a_i|a_{i+1}$. In the product, $x_\infty$ makes sense, and $f(x_\infty) = f(x_n)$ modulo $a_{n+1}$. By choosing the $a_i$ inductively in a suitable way, one may derive a contradiction (for example by arranging that the absolute value of a minimal representative of $f(x_n)$ mod $a_{n+1}$ goes to $\infty$, hence can't be given by a single integer $f(x_\infty)$)

EDIT: In an earlier version, I claimed erroneously that the following statement holds for general $I$, and can be formally reduced to the case of countable $I$. I currently do not see how to perform that reduction. The argument also had some gaps, which I hopefully fixed now. So here's an argument for countable $I$:

For countable $I$, we have $\operatorname{Hom}(\prod_I \mathbb{Z}, \mathbb{Z})\cong \bigoplus_I \mathbb{Z}$. The key statement is that a homomorphism $\prod_I \mathbb{Z} \to \mathbb{Z} $ necessarily has "finite support", i.e. there is a finite subset $I'\subseteq I$ such that $f$ factors through the projection $\prod_I \mathbb{Z}\to \prod_{I'}\mathbb{Z}$.

If that is not the case, then (identifying $I$ with $\mathbb{N}$) for any $n$, we find $v_n\in \prod_{\mathbb{N}} \mathbb{Z}$ with $v_n(i)=0$ for $i<n$ and $f(v_n)\neq 0$. But then we may consider sequences $a_i$ with $a_i\mid a_{i+1}$, and form $x_n = \sum_{i=0}^n a_i v_i$. Note this also makes sense for $n=\infty$. Now observe that $f(x_n) = f(x_\infty)$ modulo $a_{n+1}$, but if we inductively fix the elements $a_n$, we may arrange for the absolute value of the minimal representative of $f(x_n)$ modulo $a_{n+1}$ to diverge, leading to a contradiction. Simply pick $a_n$ large enough so that $f(x_n)$ is larger than $f(x_i)$ for all $i<n$, and then ensure that additionally $|a_{n+1}| > 2|f(x_n)|$ when we pick $a_{n+1}$.

Why does this help? If $M$ is a retract of $\prod_I \mathbb{Z}$, then $\operatorname{Hom}(M,\mathbb{Z})$ is a retract of $\bigoplus_I \mathbb{Z}$, hence free, and so $\operatorname{Hom}(\operatorname{Hom}(M,\mathbb{Z}),\mathbb{Z})$ is a product of copies of $\mathbb{Z}$. In general, this does not agree with $M$, however, in this situation it does! We always have a natural transformation $M\to \operatorname{Hom}(\operatorname{Hom}(M,\mathbb{Z}),\mathbb{Z})$, let us call an abelian group "reflexive" if this is an isomorphism. The above argument shows that $\prod_I \mathbb{Z}$ is reflexive, and since retracts of isomorphisms are retracts, it follows more generally that every retract of $\prod_I \mathbb{Z}$ is reflexive.