The answer is yes with $k = 2d + 1$. (Can this be improved?)

It is a by-product of the so-called Reidemester-Scheier method [1, Proposition II.4.1].

Claim. Let $G$ be a group and let $H$ be a subgroup of $G$ with index $d$ in $G$. Let $S$ be a generating subset of $G$. Then $H$ can be generated by elements of length at most $k = 2d + 1$.

Proof. Using the notation of [1, Proposition II.4.1], we consider a group presentation $F/N$ of $G$ with $F$ a free group with basis $X$, $\tilde{H} \subseteq F$ the preimage of $H$ in $F$, $T$ be a Schreier transersal of $\tilde{H}$ in $F$. Then $H$ is generated by elements of the form $\gamma(x, t) = tx (\overline{tx})^{-1}$ with $x \in X$, where $\overline{w} \in T$ is defined through $\tilde{H}w = \tilde{H}\overline{w}$ for $w \in F$. The result follows since every word in $T$ has length at most $d$ with respect to $X$.

[1] R. Lyndon and P. Schupp, "Combinatorial Group Theory", 1977.