It works for any ring in which at most one prime is not invertible (in particular it works for any field). In other words, if there is a ring homomorphism $\mathbb{Z}_{(p)} \rightarrow R$ for some prime $p$ (where $\mathbb{Z}_{(p)}$ is the ring of $p$-local integers), then $\mathbb{Z}$ can be replaced with $R$ in Rim's theorem.
The reason is the following: Let $H \leq G$ and consider the restriction functor $\text{Res}_H^G: RG\text{-Mod} \rightarrow RH\text{-Mod}$ and its left adjoint $\text{Ind}_H^G: RH\text{-Mod} \rightarrow RG\text{-Mod} .$ Then we have
Proposition: If $|G:H|$ is invertible in $R$, then the counit of the adjunction $\text{Ind}_H^G \dashv \text{Res}_H^G$ has a right inverse.
Proof. Let $E$ denote the set of left cosets of $H$ in $G$. For every $RG$-module M, define $\alpha_M: M \rightarrow \text{Ind}_H^G\text{Res}_H^G(M) = RG \otimes_{RH} M$ by $\alpha_M(m) = \frac{1}{|G:H|}\sum_{gH \in E} g \otimes g^{-1}m$ From here it is straightforward to check that $\alpha_M$'s are well-defined, $RG$-linear, natural in $M$ and the resulting natural transformation $\alpha$ is a right inverse to the counit $\varepsilon: \text{Ind}_H^G\text{Res}_H^G \rightarrow 1_{RG\text{-Mod}}$.
Corollary: If $|G:H|$ is invertible in $R$, then an $RG$-module $M$ is projective if and only if $\text{Res}_H^G(M)$ is projective.
Proof. Only if direction is clear. If $\text{Res}_H^G(M)$ is projective, so is the induced module $\text{Ind}_H^G\text{Res}_H^G(M)$, which has a direct summand isomorphic to $M$ by the above proposition. Hence $M$ is projective.
Now if every prime other than $p$ is invertible in $R$, let $H \in \text{Syl}_p(G)$ and apply the above corollary; we get the analogue of Rim's theorem for $R$.