Yes, because $SU_3(q) \subset SU_3(q^2)$ is the subgroup of elements Galois-conjugate to their inverse. $SU_3(q')$ is the subgroup of elements defined over $q'$ and Galois-conjugate to their inverse. Since the relevant Galois action is the same, they are the same. Then we mod out by the centers, which preserves inclusions.

Yes, because $SU_3(q) \subset SU_3(q^2)$ is the subgroup of elements Galois-conjugate to their inverse. $SU_3(p)$ is the subgroup of elements defined over $p$ and Galois-conjugate to their inverse. Since the relevant Galois action is the same, they are the same. Then we mod out by the centers, which preserves inclusions.