Pete's reference and other remarks in his comments are absolutely the right place to look/thing to think about. Let me make a slightly more elaborate remark, which may be helpful.

A key fact which helps one think about this sort of question is the following: if
$V$ is a rep. of $G$ over $k$, and $l$ is a finite extension of $k$,
then $End_{l[G]}(l\otimes_k V) = l\otimes_k End_{k[G]}(V).$ (This is easily checked.)

At least if $k$ is perfect, then when $V$ is irreducible over $k$ the base-change
$l\otimes_k V$ will be semisimple over $l$ (i.e. a direct sum of irreds.).
Given this, one can determine its structure (is it a sum of distinct irreds., say, or does it contain two copies of the same irrep.?) by looking at $End_{l[G]}(l\otimes_k V)$, which
as I already noted we can compute as $l\otimes_k End_{k[G]}(V)$.

Let's suppose that $l = \bar{k}$, since that's probably the case of greatest interest.
Then if $W$ is a direct sum of mutually non-isomorphic irreps., then $End_{\bar{k}[G]}(W)$
is a product of copies of $\bar{k}$, as many as there are summands of $W$.
(If $W = \oplus_i W_i,$ then we get one copy of $\bar{k}$ for each $W_i$, since $End_{\bar{k}[G]}(W) = \bar{k}$, but there are no maps between the different $W_i$,
since they are non-isomorphic by assumption.) On the other hand, if say $W$ were
a direct sum $W = W_1\oplus W_2,$ then $End_{\bar{k}[G]}(W) = M_2(\bar{k})$
(since $Hom_{\bar{k}[G]}(W_i,W_j) = \bar{k}$ for any choice of $i$ and $j$).

Now using our formula $End_{\bar{k}[G]}(\bar{k}\otimes_k V) = \bar{k}\otimes_k End_{k[G]}(V),$ we see that $\bar{k}[G]\otimes_k V$ is a direct sum of *distinct*
irreps. if and only if $End_{k[G]}(V)$ is a field (since it is precisely this case which
gives a product of copies of $\bar{k}$ when we tensor up with $\bar{k}$), while
$\bar{k}\otimes_k V$ will contain *multiple copies* of some irrep. if and only
if $End_{k[G]}(V)$ is a non-commutative division ring, since then extending scalars to
$\bar{k}$ will give a non-trivial matrix ring over $\bar{k}$. (This explains Pete's remark about the quaternion group in his comment above.)
More precisely, we see that if $End_{k[G]}(V)$ is a division ring of dimension $n^2$
over its centre, which has say degree $d$ over $k$, then
$\bar{k}\otimes_k V$ will be a product of $n$ copies each of $d$ distinct irreps. over $\bar{k}$.

Linear Representations of Finite Groupsfor a nice introduction to representation theory over a nonalgebraically closed field of characteristic $0$. Some keywords: Schur indices, Schur subgroup. M. Isaacs's book on representation theory covers these topics in somewhat greater depth. – Pete L. Clark Jul 18 '10 at 18:57Algebra). So the centre of any division ring that arises as the endomorphism ring of an irrep over a characteristic zero field $k$ will be an abelian extension of $k$. But over $\mathbb{Q}$ or $\mathbb{Q}_p$ the only examples of noncommutative endomorphism rings of irreps I've seen have been quaternionic. Are there examples where the degree of the division ring over its centre is greater than $4$? – Robin Chapman Jul 18 '10 at 20:38