In general $Ext^1_R(R/I,M) \not\cong M/IM$. 

As an example take a finite abelian group $G$ and set $R := \mathbb{Z}G$ and let $I:= I_G = \ker(\mathbb{Z}G \to \mathbb{Z},\; g \mapsto 1)$ be the augmentation ideal. $I_G$ is a prime ideal since $\mathbb{Z}G/I_G \cong \mathbb{Z}$. Then, with trivial coefficients 
$$Ext_R^1(R/I,\mathbb{Z})=Ext_{ZG}^1(\mathbb{Z},\mathbb{Z})=H^1(G;\mathbb{Z})=Hom(G,\mathbb{Z})=0$$
while $\mathbb{Z}/I_G\mathbb{Z}=\mathbb{Z}$. The latter holds because $I_G$ is generated by $g-1$ $(g \in G)$ and $(g-1) \cdot 1 = 0$. 

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In general, the following holds:
$$Ext^1_R(R/I,M)=\dfrac{Hom_R(I,M)}{i(M)}$$
where $i: M \to Hom_R(I,M),\; m \mapsto (x \mapsto xm)$. But it seems hard to get a closed expression for $Hom_R(I,M)$ if $I$ is not free as $R$-module.