The question gets a little confusing when algebraic group language is added. But on the level of finite groups of Lie type, it's useful to keep in mind that Jordan decomposition in the algebraic group sense has intrinsic meaning: given $p$, 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.
In any case, the initial question concerns only finite groups. 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 $\mathbb{F}_q$). So the only way a finite group representation could map a diagonal matrix to one which is not diagonalizable over $k$ is for $k$ to have fewer than $q$ elements. But $G$ is simple if $p=2$ and simple modulo a center of order 2 if $p$ is odd, while it has an element of order $q-1$, so such a representation can't exist.