OK, I think I have an example of two groups with the same profinitization and a computable property which distinguishes them. The point is that very fine detail about the commutator subgroups can't be seen in the profinitization.

Let $q$ be prime and let $K$ be the $q$-th cyclotomic field.
Choose $q$ such that the class group of $K$ is not trivial. Let $I$ be a trivial ideal of $\mathcal{O}_K$ and $J$ a nontrivial ideal. Our groups $G$ and $H$ will be $(\mathbb{Z}/q) \ltimes I$ and $(\mathbb{Z}/q) \ltimes J$.

For any group $B$, let $B' = [B,B]$ and $B'' = [B', B']$. Note that $B/B'$ acts on $B'/B''$ by conjugation. Our computable criterion is the following:

$B/B' \cong \mathbb{Z}/q \times \mathbb{Z}/q =: A$, the action of the group ring $\mathbb{Z}[A]$ on $B'/B''$ factors through a map $\mathbb{Z}[A] \to \mathcal{O}_K$ and, as such, $B'/B''$ is a free $\mathcal{O}_K$ module.

We leave it as an exercise that $G$ satisfies this condition and $H$ does not.

I believe this condition should be computable. ~~We can go from a finite presentation of $B$ to one of $B'$.~~ (**UPDATE** I have revised this argument.) Abelianizations are computable, so we can check whether $B/B'$ has the right format. If it does, then $B'$ has finite index in $B$. I think we can use this to get a finite presentation of $B'$: Let $\Delta$ be a two-dimensional $CW$-complex with one vertex, an edge for each generator of $B$ and a two cell for each relation. Let $\Delta'$ be the cover of $B$ corresponding to $B'$. Since $B$ has finite index in $B'$, $\Delta'$ will have finitely many cells, and we get a finite presentation of $B'$.

We can the compute the abelianization of $B'$ and, I think, the action of the abelianization of $B$ on that of $B'$ should be computable. Note that there are only $q^2$ maps from $\mathbb{Z}[A]$ to $\mathcal{O}_K$, so we can just check them each in turn. The class of a finite generated module for a Dedekind domain should be computable by standard number theory methods, although I admit I couldn't describe them.

The fact that these two groups have the same profinitization is relatively well known. Let $\hat{I}$ and $\hat{J}$ denote the profinite completions of $I$ and $J$. The profinite completions of $G$ and $H$ are $\mathbb{Z}/n \ltimes \hat{I}$ and $\mathbb{Z}/n \ltimes \hat{J}$.

We can identify $\hat{I}$ and $\hat{J}$ with submodules of $\mathbb{A}^0_K$, the integral adeles of $K$. Since $I$ and $J$ are locally principal, these are principal ideals in the ring $\mathbb{A}^0_K$. They are thus equivalent as $\mathbb{A}^0_K$ modules, and thus as $\mathcal{O}_K$ modules.