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.