Yes, because $G$ is top. f.g. and its maximal subgroups have bounded index.

Let me prove the latter fact. Let $K$ be the core (=intersection of conjugates) of $H$, so $K$ is an open normal pro-$p$-subgroup. Let $M$ be a maximal open subgroup in $G$. Then either $MK=M$ or $MK=G$. In the first case, $K\subset M$, which bounds the index of $M$.

So now assume $MK=G$. Let $N$ be the core of $M$. Write $G'=G/N$ and also denote with primes the images of the various subgroups in $G'$; since $N\subset M$ we have $M'\neq G'$, so $M'$ is maximal in $G'$. Also $G'=M'K'$, so necessarily $K'$ is nontrivial. Let $A$ be the (nontrivial) center of the $p$-group $K'$; it is characteristic in the normal subgroup $K'$ and hence is normal in $G'$. Then $A\cap M'$ is normalized by both $M'$ and by $N'$, hence it is normal in $G'$. Since the core of $M'$ is trivial, we deduce that $A\cap M'=1$. By maximality of $M'$, we deduce that $G'=M'\ltimes A$. The centralizer of $A$ in $M'$ is normal in $G'$ and again since the core of $M'$ is trivial it has to be trivial. Thus $K'\cap M'=1$, and since $K'$ contains $A$ this forces $K'=A$. Thus
$$|M'|=[G':K']\le [G:K] $$
Hence $M'$ has bounded cardinal. Since by maximality $M'$ acts irreducibly on $A$ (in the sense that $A\neq 0$ and $A$ has no nontrivial $M'$-invariant subgroup), $A$ is $p$-elementary, and has dimension $\le |M'|$, hence cardinal dividing $p^{[G:K]}$. Since the cardinal of $A$ equals the index of $M'$, we deduce that $[G:M]=[G':M']$ divides $p^{[G:K]}$, so is bounded.