The literature about constructing model structures on abelian categories has grown significantly since Hovey's book came out. In particular, there is now a connection between this process and cotorsion pairs which Hovey found in a series of papers from around 2002-2004 and which his student Jim Gillepsie fleshed out in subsequent years. Here is one of Hovey's papers (as a pdf). If you google cotorsion pairs you should find more. Ed Enochs and Sergio Estrada also have several good papers on this. Another MO question which elaborates on this topic is here.

An abelian model category is a category which is abelian and has a model structure, along with a compatibility condition between those two things. So in particular, you need the cofibrations to be monomorphisms with cofibrant cokernel and the fibrations to be epimorphisms with fibrant kernel. When you have an abelian model category (i.e. abelian category with a model structure and a compatibility condition) then you get two cotorsion pairs from it. Conversely, given a cotorsion pair you can construct a model structure if your cotorsion pair is nice (it has to be "complete" which means you need to have enough projectives and enough injectives).

In your case, $R$-mod has two complete cotorsion pairs: $(C, W\cap F)$ and $(C\cap W, F)$, by Prop 2.2. in the linked paper. You can promote those to pairs on $I^{R-mod}$ via $(I^C, I^{W\cap F})$ and $(I^{C\cap W}, I^F)$ using the fact that Ext groups between diagrams should be computed object-wise (I can't imagine how else you would compute them). It's also easy to see that a diagram of elements of the form $W\cap F$ is an intersection of diagrams from $I^W$ and $I^F$, so you can write $I^{W\cap F}$ as $I^W \cap I^F$. You say in your question that the diagrams have enough projectives and injectives, so this means you get complete cotorsion pairs. All you have to do to use Theorem 2.5 of the paper and hence get an abelian model structure is to show $I^W$ is thick, i.e. closed under retracts and satisfies the 2-out-of-3 property. Both should be easy, since $W$ is thick.