Let me add a useful reference book, probably no longer in print but found in many libraries: R.W. Carter, Finite Groups of Lie Type: Conjugacy Classes and Complex Characters, Wiley-Interscience, 1985. The book includes a lot of information about the irreducible representations of Weyl groups (though not with complete proofs) in the context of characters of finite groups of Lie type, unipotent classes in simple algebraic groups, Springer correspondence, etc. Chapters 11 and 13 involve the Weyl groups most heavily. (Though the book by Chriss and Ginzburg goes much deeper into the geometry of the Springer correspondence, the treatment there is mostly limited to the case of symmetric groups which is often more straightforward than the general case; for this Carter gives a helpful overview.)
For example, the $25 \times 25$ character table of the Weyl group of type $F_4$ originally worked out by T. Kondo in a 1965 journal article is displayed on page 413 of the book, while the role of these characters in the Springer correspondence is summarized on page 428 (involving the previous study of unipotent classes for the simple algebraic group of type $F_4$). Like all character tables of Weyl groups, this one has entries in $\mathbb{Z}$. It's not at all easy in a case like $F_4$ to write down explicit integral matrices affording the irreducible representations, but fortunately the characters alone are sufficient for some applications like those developed by Carter. One caveat is that notation for conjugacy classes and characters differs in various sources.
Short answer: you don't want to consider group cohomology as defined for finite groups for Lie groups like $U(1)$, or indeed topological groups in general. There are other cohomology theories (not Stasheff's) that are the 'right' cohomology groups, in that there are the right isomorphisms in low dimensions with various other things.
Long answer:
Group cohology, as one comes across it in e.g. Ken Brown's book (or see these notes), is all about discrete groups. The definition of group cocycles in $H^n(G,A)$, for $A$ and abelian group, can be seen to be the same as maps of simplicial sets $N\mathbf{B}G \to \mathbf{K}(A,n)$, where $\mathbf{B}G$ is the groupoid with one object and arrow set $G$, $N$ denotes its nerve and $\mathbf{K}(A,n)$ is the simplicial set corresponding (under the Dold-Kan correspondence) to the chain complex $\ldots \to 1 \to A \to 1 \to \ldots$, where $A$ is in position $n$, and all other groups are trivial. Coboundaries between cocycles are just homotopies between maps of simplicial spaces.
The relation between $N\mathbf{B}G$ and $K(G,1)$ is that the latter is the geometric realisation of the former, and the geometric realisation of $\mathbf{K}(A,n)$ is an Eilenberg-MacLane space $K(A,n)$, which represents ordinary cohomology ($H^n(X,A) \simeq [X,K(A,n)]$, where $[-,-]$ denotes homotopy classes of maps). This boils down to the fact that simplicial sets and topological spaces encode the same homotopical information. It helps that $N\mathbf{B}G$ is a Kan complex, and so the naive homotopy classes are the right homotopy classes, and so we have
$$sSet(N\mathbf{B}G,\mathbf{K}(A,n))/homotopy \simeq Top(BG,K(A,n))/homotopy = [BG,K(A,n)]$$
In fact this isomorphism is a homotopy equivalence of the full hom-spaces, not just up to homotopy.
If we write down the same definition of cocycles with a topological group $G$, then this gives the 'wrong' cohomology. In particular, we should have the interpretation of $H^2(G,A)$ as isomorphic to the set of (equivalence classes of) extensions of $G$ by $A$, as for discrete groups. However, we only get semi-direct products of topological groups this way, whereas there are extensions of topological groups which are not semi-direct products - they are non-trivial principal bundles as well as being group extensions. Consider for example $\mathbb{Z}/2 \to SU(2) \to SO(3)$.
The reason for this is that when dealing with maps between simplicial spaces, as $N\mathbf{B}G$ and $\mathbf{K}(A,n)$ become when dealing with topological groups, it is not enough to just consider maps of simplicial spaces; one must localise the category of simplicial spaces, that is add formal inverses of certain maps. This is because ordinary maps of simplicial spaces are not enough to calculate the space of maps as before. We still have $BG$ as the geometric realisation of the nerve of $\mathbf{B}G$, and so one definition of the cohomology of the topological group $G$ with values in the discrete group $A$ is to consider the ordinary cohomology $H^n(BG,A) = [BG,K(A,n)]$.
However, if $A$ is also a non-discrete topological group, this is not really enough, because to define cohomology of a space with values in a non-discrete group, you should be looking at sheaf cohomology, where the values are taken in the sheaf of groups associated to $A$. For discrete groups $A$ this gives the same result as cohomology defined in the 'usual way' (say by using Eilenberg-MacLane spaces).
So the story is a little more complicated than you supposed. The 'proper ' way to define cohomology for topological groups, with values in an abelian topological group (at least with some mild niceness assumptions on our groups) was given by Segal in
G Segal, Cohomology of topological groups, in: "Symposia Mathematica, Vol. IV (INDAM, Rome, 1968/69)", Academic Press (1970) 377–387
and later rediscovered by Brylinski (it is difficult to find a copy of Segal's article) in the context of Lie groups in this article.
Best Answer
You have already used the following fact in your statement:
For every representation $\rho: G\to \mathrm{Aut}_R(A)$ ($A$ an $R$-module) we can give $A$ the structure of an $R[G]$-module by $g\cdot a = \rho_g(a)$. Conversely, any torsion free $R[G]$-module $A$ admits a representation $\rho: G \to \mathrm{ Aut}_R(A)$ given by $\rho_g(a) := g\cdot a$.
So studying $R[G]$-modules IS studying representations of $G$. My answer would be if you want to study representations study $R[G]$-modules. $R[G]$-modules aren't like a cure-all, because they don't give you an action to start with. Does anyone know how to classify actions?
I'm not sure if my answer was helpful. Hopefully it saved you some time. I don't think I understood what you meant about "taking the fixed elements to get the cohomology".