As mentioned in the other answer, you cannot prove in $\sf ZF+\lnot AC$ that $\Bbb R$ and $\Bbb C$ are isomorphic as additive groups.
On the one hand, the failure of choice is not a localized statement. It tells you nothing about where it fails, it is consistent that any "reasonable set" and its power set are all well-orderable, so all the theorems with the axiom of choice follow immediately. In particular, it is perfectly reasonable that choice fails but $\Bbb R$ can be well-ordered, in which case there are Hamel bases for $\Bbb R$ and $\Bbb C$ over $\Bbb Q$, and the groups are isomorphic.
On the other hand, assumptions such as "Every set of reals has the Baire property" , "Every set of reals is Lebesgue measurable", or The Axiom of Determinacy (which itself implies the two former statements) imply—at least in the presence of Dependent Choice—that any homomorphism between Polish groups is continuous, so in particular any homomorphism between $\Bbb R$ and $\Bbb C$ is continuous. But it is fairly easy to show that no isomorphism can be continuous.
One should remark, however, about the consistency strength of such failures. If all sets are Lebesgue measurable (again, assuming Dependent Choice), then a theory stronger than just $\sf ZF$ is consistent. Even more in the case of Determinacy. However, the assumption that every set has the Baire property does not increase the consistency strength.