Write $G=F/N$, where $F$ is free over $S$. Let $C_F$ and $C_G$ be the Cayley graphs of $F$ and $G$: then $C_G=C_F/N$ and in particular the fundamental group of $C_G$ is naturally $N$ (note that we need the convention that if $1\in S$ then it yields a self-loop and if $S$ has elements of order 2 then they yield double edges; if this is not fine, we can assume that $S$ has no element of order $\le 2$). Also $H_1(C_G)$ is $N/[N,N]$.

The condition that $R\subset F_S$ generates $H_1(C_G)$ means that the subgroup $L$ generated by $R$ satisfies $L[N,N]=N$. The condition that $\langle S\mid R\rangle$ is a presentation means that the subgroup of $F_S$ normally generated by $L$ is equal to $N$.

Assume that $M\subset F_S$ is a normal subgroup and is contained in $F_N$, and that $N/M$ is a perfect group. Then the subgroup normally generated by $M$ is just $M$ itself. The condition that $N/M$ is perfect means that $[N,N]M=N$. So to show that the answer is no, it is enough to pick $R=M$ and assume $N/M\neq 1$.

So the recipe is just to pick $K$ a nontrivial perfect finitely generated group, $S$ a finite symmetric generating subset (say, with no element of order $\le 2$), $G$ a cyclic group (which can be arranged to be of order 5, if $S$ has 4 elements) with an injective map $i:S\to G$ with generating image (again denoted by $S$, say with no element of order $\le 2$); let $\bar{S}\subset S\times G$ be its graph: this is a generating subset. So we pick $F_S/M$ to be the direct product $S\times G$ (mapping $S$ to $\bar{S}$ in the canonical way) and $F_S/N$ to be $G$ (mapping $S$ to $G$ using $i$).

I leave the explicit computations, which seem doable if one starts with a presentation of a nontrivial perfect group (they yield $R$ infinite but some finite subset is enough when $G$ is finite).