The answer is yes.
Under the assumption that $H$ is normal in $G$, an upper bound on the length of generators of $H$ with respect to $S$ is given by $k(d) := d^2 - d + 1$. If $G$ is finitely generated and $H$ is an arbitrary subgroup of index $d$ in $G$, then $k(d!)$ is an upper bound.
This follows from a classical combinatorial result of Schreier [1, Proof of Proposition I.3.7].
These ideas of Schreier go back to 1927 and already establish [2, Lemma 3.4] of Shalen and Weigreich (1992), that is, they show that $2d - 1$ is upper bound when $S = S^{-1}$.
For $S$ a subset of a group $G$ and $n \in \mathbb{N}_{\ge 0}$, we denote by $S^{\le n}$ the set of elements of $G$ which can be written as product of at most $n$ elements of $S$. The monoid $S^{\ge 0} := \bigcup_{n \ge 0}S^{ \le n}$ is the submonoid of $G$ generated by $S$.
Example. Let $G$ be a finite group. Then $G = S^{\ge 0}$ for every generating subset $S$ of $G$.
Claim 1. Let $G$ be a group and let $H$ be a normal subgroup of finite index $d$ in $G$. Let $S$ be a generating subset of $G$. Then $H$ is generated by some subset of $S^{\le k(d)}$.
Claim 2. Let $G$ be a finitely generated group and let $H$ be a subgroup of finite index $d$ in $G$. Let $S$ be a generating subset of $G$. Then $H$ is generated by some subset of $S^{\le k(d!)}$.
We will use the following lemma.
Lemma 1. Let $G$ be a finite group. Then $G = S^{\le \vert G \vert - 1}$ for every generating subset $S$ of $G$.
Proof of Lemma 1. In the connected Cayley graph of the monoid $G$ with respect to $S$, every path of length at least $\vert G \vert$ which connects $1$ to $g \in G$ contains a loop. As loops represent the trivial element of $G$, any such path can be turned into a shorter path connecting $1$ to $g$.
Definition 1. Let $G$ be a group and le $H$ be a subgroup of $G$. A transversal of $H$ in $G$ is a subset $T$ of $G$ such that, for distinct $t \in T$, the cosets $Ht$ are distinct and $G$ is the union of the cosets $Ht$.
Lemma 2. [1, Proof of Proposition I.3.7] Let $G$ be a group and let $H$ be a subgroup of $G$. Let $T$ be a transversal of $H$ in $G$ containing $1$ and let $S$ be a generating subset of $G$. For $g \in G$, define $\overline{g} \in T$ through $H\overline{g} = Hg$ and set $\gamma(g) = g \overline{g}^{-1}$. Then $H$ is generated as a group by the elements $\gamma(ts)$ with $t \in T, s \in S$.
Proof of Lemma 2. Since $T$ is a transversal of $H$ in $G$ which contains $1$, we have $H = \{\gamma(g) \, \vert \, g \in G\}$. Using the identity $\overline{\overline{g}s} = \overline{gs}$, we obtain that $\gamma(gs) = \gamma(g)\gamma(\overline{g}s)$ for every $g \in G, s \in S$. Using the identity $\overline{\overline{ts}s^{-1}} = t$, we also obtain that $\gamma(ts)^{-1} = \gamma(\overline{ts}s^{-1})$. Reasoning by induction on the length of a word on $S \cup S^{-1}$ representing $g \in G$, we infer that $H$ is generated by the elements $\gamma(ts)$.
Proof of Claim 1. Applying Lemma 1 to $G/H$ and the image of $S$ in $G/H$, we find a transversal $T$ of $H$ in $G$ such that $1 \in T \subseteq S^{\le d - 1}$.
By Lemma 2, the subgroup $H$ is generated (as a group) by elements of the form $\gamma(ts)$ with $s \in S, t \in T$. Since $H$ is also generated as a group by the elements of the form $\overline{ts}^d, \gamma(ts)\overline{ts}^d = ts \overline{ts}^{d - 1}$, the result follows.
In order to prove Claim 2, we shall resort to the normal core $\operatorname{core}_G(H) = \bigcap_{g \in G}g^{-1}Hg$ of $H$ in $G$.
We assume from now on that $G$ is finitely generated. If $H$ is of index $[G:H] = d$ in $G$, then $[G: \operatorname{core}_G(H)] \le d!$, see [1, Theorem IV.4.7] for ideas as how to prove the latter inequality.
Proof of Claim 2. By Lemma 1, applied to $H/\operatorname{core}_G(H)$, there is a transversal $T$ of $\operatorname{core}_G(H)$ in $H$ such that $1 \in T \subseteq S^{\le d! - 1}$. As $H$ is generated by $T$ together with $\operatorname{core}_G(H)$, the result follows from Claim 1 applied to $G$ and $\operatorname{core}_G(H)$.
A remark on non-negative Schreier transversals.
We can relate the proofs of Claims 1 and 2 to the topological approach of the proof of [2, Lemma 3.4]. The approach of P. Shalen and P. Wagreich relies on a spanning subtree of some $d$-sheeted covering graph of a bouquet of circles labelled by the elements of $S$. The combinatorial analog of this subtree is a Schreier transversal.
Definition 2. Let $G$ be a free group with basis $S$. A Schreier transversal of a subgroup $H$ in $G$ is a transversal $T$ of $H$ in $G$ such that every initial segment of an element $t \in T$, as a reduced word over $S \cup S^{-1}$, is in $T$.
We can now rephrase Claim 1 and its proof in the following way.
Lemma 3. Let $G$ be a free group with basis $S$. Let $H$ be a normal subgroup of finite index $d$ in $G$. Then there is a Schreier transversal of $H$ in $G$ that is contained in $S^{\le d - 1}$.
Proof of Lemma 3. We call $T$ a partial non-negative Schreier transversal of $H$ in $G$, if $T \subseteq S^{\ge 0}$, $Ht \neq Ht'$ whenever $t, t' \in T$ are distinct and if every initial segment of an element $t \in T$, as a reduced word over $S$, is in $T$. Let $T_0$ be maximal partial non-negative Schreier transversal of $H$ in $G$. The maximality of $T_0$ implies that $Hts \subseteq HT_0$ for every $t \in T_0$ and $s \in S$. Thus $HS^{\ge 0} \subseteq HT_0$. Since $G = HS^{\ge 0}$, the set $T_0$ is a Schreier transversal.
[1] R. Lyndon and P. Schupp, "Combinatorial Group Theory", 1977.
[2] P. Shalen and P. Wagreich, "Growth Rates, Zp-Homology, and Volumes of Hyperbolic 3-Manifolds", 1992
