Do all the homomorphisms $\phi: SL(2,\mathbb{Z})\ltimes \mathbb{Z}^2 \to GL(2,\mathbb{R})$ always have that $\phi_{\mathbb{Z}^2}$ is trivial, i.e. $\phi(\mathbb{Z}^2)=I_2$?

Yes. Consider $\mathbb R^2$ as a representation of the semidirect product, hence as a representation of $\mathbb Z^2$, and write it as an extension of irreducible characters $\chi_1$, $\chi_2$. If any nontrivial character $\chi$ appears, then all conjugates of that character by automorphisms of $\mathbb Z^2$ in $SL_2(\mathbb Z)$ must appear as well. But each nontrivial character has at least three distinct conjugates (with the minimum achieved by the three nontrivial quadratic characters), so none can appear. Hence the representation of $\mathbb Z^2$ is unipotent, but this just gives a character $\mathbb Z^2 \to \mathbb R^{+}$. Again all conjugates must appear, but there is just one conjugate, so it must be trivial. So $\mathbb Z^2$ acts trivially. 

