Because only infinite dimensional, unitary representation of $SL(2,F)$ for a local field $F$ can fail to be tempered, if they are spherical. This follows from the classification. This is true for $GL(2)$ as well modulo twisting by one-dimensional representations, which are trivial for $SL(2)$.

As Asaf points out, looking at the K=SU(2)-invariant vectors will do the job. You will obtain Eisenstein series and cusp forms. There you actually have non-tempered representations besides the trivial representation. There are atmost finitely many. For $\Gamma$ a congruence subgroup of an imaginary quadratic field $k$, it is an important conjecture that there are non (the analogue of the Selberg eigenvalue conjecture). Then you actually have a much bigger group acting, i.e., $SL_2(A_k)$ via strong approximation, i.e. there exists an open subgroup$K_\Gamma$ of $SL_2(A_{k,f})$ (finite adeles) 
$$ \Gamma \backslash SL_2(\mathbb{C}) = SL_2(k) \backslash SL_2(A_k) / K_\Gamma.$$

If you consider an irreducible representation in there, it factors into representations of $SL_2(k_v)$ for each place. It is assumed that they should all be tempered (besides the one-dimensional representations). This is know as the Ramanujan Peterson conjecture. Non trivial bounds are known due to Blomer and Brumley (they actually work with $GL(2)$).