I think the answer is yes if and only if $k$ contains all $(q-1)$-st roots of unity, for the following reason:

If $D$ is a diagonal matrix in $SL_r(\mathbb F_q)$, then its order divides $q-1$ because all entries on the diagonal are in $\mathbb F_q^\ast$. But every endomorphism of $k^n$ of finite order $n$ prime to $p$, and in particular $\varphi(D)$, is diagonalisable if $k$ contains all $n$--th roots of unity.

On the other hand: Take $D$ of order $q-1$ and suppose $\varphi(D)$ is diagonalisable over $k$. Then all eigenvalues of $\varphi(D)$ are in $k$. These eigenvalues are roots of unity, such that the lcm of their orders is $q-1$. This implies that $k$ contains primitive $(q-1)$-st roots of unity.

As usual I'm posting in a hurry and have not checked anything.

I think the answer is yes if and only if $k$ contains all $(q-1)$-st roots of unity, for the following reason:

If $D$ is a diagonal matrix in $SL_r(\mathbb F_q)$, then its order divides $q-1$ because all entries on the diagonal are in $\mathbb F_q^\ast$. But every endomorphism of $k^n$ of finite order $n$ prime to $p$, and in particular $\varphi(D)$, is diagonalisable if $k$ contains all $n$--th roots of unity.

On the other hand: Take $D$ of order $q-1$ and suppose $\varphi(D)$ is diagonalisable over $k$. Then all eigenvalues of $\varphi(D)$ are in $k$. These eigenvalues are roots of unity, such that the lcm of their orders is $q-1$. This implies that $k$ contains primitive $(q-1)$-st roots of unity.

As usual I'm posting in a hurry and have not checked anything.

Edit: Just to make the statement clear: Given $p, q$ and $k$ as in the question, I claim the followimg to be equivalent:

(a) the field $k$ contains all $(q-1)$-th roots of unity
(b) for every representation $\varphi: SL_2(\mathbb F_q) \to GL_n(k)$ and every diagonal matrix $D\in SL_2(\mathbb F_q)$, the matrix $\varphi(D)$ is diagonalisable over $k$.