This post came up on my RSS feed, and I didn't see that it had been answered in 2011, but since I've now written the following, you may as well have it ...
One other point, which I think is relevant to your question. If you take $G$ over $k=\bar k$, char $k=p$ with a Frobenius map $F$ whose fixed points $G^F$ is a split Chevalley group $G^F=G(q)$ with $q=p^r$, then CPSvdK in their 1977 Inventiones paper explain a strong connection between the rational cohomology $H^i(G,M)$ and $H^i(G(q),M)$ for $M$ a finite-dimensional rational $G$-module. As other posters have pointed out, there is divergence between the two, even for $q$ sufficiently large. One has to perform sufficiently many Frobenius twists to $M$ to make the two sides agree. The first example is for $G=SL_2$ and $k$ has characteristic $2$. Then if the natural module is $V=L(1)$, this has $H^1(SL_2,V)=0$ in the rational setting but for all $r>1$, $H^1(SL_2(2^r),V)$ is $1$-dimensional. The 'generic cohomology' is achieved as the stable limit $H^i(G(q),M)$ for $q=p^r$ as $r\to \infty$. So $H_\mathrm{gen}^1(SL_2,V)\cong k$.
What's the difference between $H^i$ and $H^i_\mathrm{gen}$? Well, with a little computation you can see that in order to get the non-trivial cocycles of $H^1_\mathrm{gen}(SL_2,V)$ in $H^1(SL_2,V)$ you would have to be able to take square roots. Well, of course, you can't do this in the category of algebraic morphisms.
But there is also another cohomology coming from forgetting all variety structure and allowing maps to be non-morphisms. So we're just looking at $G$ as a(n infinite) group. We could denote this $H^i_{\mathrm{abs}}(G,M)$ (where the 'abs' is for abstract). In the category of abstract $G$-modules, of course you can take any set theoretic map you like, so square roots are fine. And in fact one has an isomorphism $H^i(G_\mathrm{abs},M)\cong H^i(G_\mathrm{gen},M)$. I am led to believe that there is an article by Brian Parshall, [EDIT: with thanks to Wilberd van der Kallen] 'Cohomology of algebraic groups' in [The Arcata conference on Representations of Finite Groups, Proc Symp Pure Math 47 Part 1] which gives a proof of this for finite-dimensional rational modules $M$ using an argument he attributes to van der Kallen.