This is an edited version of an earlier hasty try, which went off the tracks and was removed. As both Xandi Tuni and Someone have made clear in their comments, the answer to the stated question is "no": even if the representation is otherwise well-behaved, it can easily take a *diagonal* matrix to one which is not diagonalizable over too small a field (but is semisimple and therefore does diagonalize over an extension field).

The question gets a little confusing when algebraic group language is added. But on the level of finite groups in nonzero characteristic, it's useful to keep in mind that *Jordan decomposition* in the algebraic group sense has intrinsic meaning: given a prime $p$, an element of a finite group decomposes uniquely as a product of two commuting elements, one of $p$-power order and the other of order prime to $p$. This decomposition is preserved under any finite group homomorphism. So for matrix groups a semisimple element is sent to a semisimple element, but a larger field may be needed to diagonalize such a matrix.

In any case, the initial question concerns only finite groups of Lie type. Just the case when the representation is *nontrivial* needs to be considered.

EDITED (with more details, as suggested in comments):

(1) For the given group $G =$ SL$_2(q)$, whose order is $q(q-1)(q+1)$, the matrices which are already diagonal have orders dividing $q-1$ (while other semisimple elements have orders dividing $q+1$ and can be diagonalized over a quadratic extension of $\mathbb{F}_q$). On the other hand, when $q>3$ the group $G$ is simple if $p=2$ and simple modulo a center of order 2 if $p$ is odd, while it has an element of order $q-1$. Suppose the representation takes such an element to a matrix which is *diagonalizable* over $k$. This matrix has order at least $(q-1)/2$, since the representation is nontrivial and $G$ is almost simple. Using the easy inequality $(p^e−1)/2 > p^{e-1}−1$, it follows that $k$ includes the $(q-1)$st roots of unity and hence includes $\mathbb{F}_q$ (with $q=p^e$).

(2) For the converse $G$ can be arbitrary. The image of an element of order dividing $q-1$ still has order dividing $q-1$. So it can be diagonalized over $k$ provided $k$ includes $\mathbb{F}_q$ (and thus contains all roots of unity of order dividing $q-1$).

All of this uses the standard characterization of subfields of a given finite field. Xandi's argument involves some of the same ideas, but fails to use the fact that the ambient group $G$ is *almost simple*. (The third paragraph isn't convincing.) For (1) it isn't enough to consider just a representation of an arbitrary finite linear group containing a diagonal matrix.