Can finite index be seen at the level of profinite completion

Let $G$ be a group, and $H$ a subgroup of $G$.

Is it possible to "see" from the profinite completions of $H$ and $G$ that $H$ has finite index in $G$?

Naively, does $H$ have finite index in $G$ iff the profinite completion of $H$ has finite index in the profinite completion of $G$? Certainly not...

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If $G$ is a group and $H$ is a finite index subgroup a result of Marshall Hall says that the profinite topology on $H$ is induced from the profinite topology on $G$. Thus the profinite completion of $H$ is the closure of $H$ in the profinite completion of $G$. Moreover, the index of the closure of $H$ in the profinite completion of $G$ is $[G:H]$. All this can be found in the book of Ribes and Zalesskii.
Note that it can happen that $H$ inherits its profinite topology from that of $G$ without being finite index, for example this happens for every finitely generated subgroup of a finitely generated free group. The closure of $H$ in the profinite completion of $G$ will then be the profinite completion of $H$ and will not have finite index.
In general the inclusion of $H$ in $G$ will not induce an inclusion of profinite completions and it is possible for a subgroup of a group to be dense in the profinite topology and be infinite index. For example let $G$ be an infinite simple group. Then the profinite completion of $G$ is trivial and so the trivial subgroup is dense.