Do groups with finitely many finite index subgroups of each index have strongly complete profinite completion? tl;dr: Let $G$ be a group with finitely many subgroups of index $n$ for each $n \in \mathbb{N}$. Is its profinite completion necessarily strongly complete?

A strongly complete profinite group is a profinite group such that all of its finite index subgroups are open. It was shown in [1] that being strongly complete is equivalent to having only finitely many finite index subgroups of index $n$ for each $n$.
It is well-known that a finitely generated group has the property of only having finitely many finite index subgroups of index $n$ for each $n$. As proven in [2], (topologically) finitely generated profinite groups are strongly complete. Thus, the profinite completion of a finitely generated group shares this property as well.
But what if $G$ has only finitely many subgroups of each index $n$, but is not finitely generated? Must its profinite completion $\hat{G}$ be strongly complete, or is there a counterexample?
[1] Smith, Wilson, On subgroups of finite index in compact Hausdorff groups.
[2] Nikolov, Segal, On finitely generated profinite groups, I: strong completeness and uniform bounds.
 A: The answer is yes for products of finite simple groups. This is proved in J. Saxl and J. S. Wilson, A note on powers in simple groups, Math. Proc. Cambridge Phil.
Soc. 122 (1997), 91–94.
I think, however, that the answer is negative. This would be based on the following:
**Fact: for every $n$ there is a finite group $H$ such that the number of squares in $H$ is $<|H|^{1/n}$ and $H$ has no proper subgroup of index $\le n$.
If the fact is proved and for each $n\ge 3$ one chooses such a finite group $H=G_n$, then $G=\prod_{n\ge 3} G_n$ works. Indeed by construction $G$ has finitely many open subgroups of each given index, and none of index $2$. But in $G_n$, some element is not a product of $n$ squares, so in each nonprincipal ultraproduct $U$ of $G$, there is a non-square. Since $U$ is then a quotient of $G$, it means that $G$ admits a quotient of order $2$.
The idea to prove the fact would be to construct perfect groups $G_n=S_n\ltimes P_n$ with $S_n$ simple of order tending to infinity, $P_n$ a 2-group, arranging that $Z(G_n)$ contains a 2-elementary subgroup $Q_n$ of order $> |G_n|^{1-1/n}$. (So the square function $G\to G$ factors through a map $G/Q_n\to G$, and hence has image of cardinal $< |G|^{1/n}$.) [I guess I already mentioned or expanded such a construction somewhere here or at MO, but right know I can't find it.]
