Let $M$ be a finitely generated $R$-module. For any filtered category $D$ any diagram $F\colon D \to R\text{-Mod}$ such that $F(f)$ is a monomorphism in $R$-Mod for all $f\in D$, it follows that the natural map $$\text{colim }\text{Hom}_R(M,F(d))\cong\text{Hom}_R(M,\text{colim }F(d))$$ is an isomorphism.
Commuting with all filtered colimits, not just those with $F(f)$ a monomorphism, should be a strictly stronger condition. But I have not been able to construct an example.
Take $R=\mathbb{Z}\left[x_{i}\right]_{\text{i}\in\mathbb{N}}$, $M=\mathbb{Z}\left[x_{i}\right]_{\text{i}\in\mathbb{N}}/\left(x_{i}\right)_{\text{i}\in\mathbb{N}}$, and consider the evident (filtered) diagram of quotients $$\mathbb{Z}\left[x_{i}\right]_{\text{i}\in\mathbb{N}}/\left(x _{i}\right)_{\text{i}\in\mathbb{N}_{<0}}\ \to\ \mathbb{Z}\left[x_{i}\right]_{\text{i}\in\mathbb{N}}/\left(x _{i}\right)_{\text{i}\in\mathbb{N}_{<1}}\ \to\ \mathbb{Z}\left[x_{i}\right]_{\text{i}\in\mathbb{N}}/\left(x _{i}\right)_{\text{i}\in\mathbb{N}_{<2}}\ \to\ \cdots$$ which is taken by $\text{Hom}_{R}\left(M,-\right)$ to the diagram $$0\ \to\ 0\ \to\ 0\ \to\ \cdots$$ but has (filtered) colimit $M$.
More generally, the compact objects of the category of modules over a commutative ring are precisely the finitely presented modules over that ring, so for every finitely-generated-but-not-finitely-presented module one can cook up such a counterexemplary diagram.