The deductive closure of $\phi$ could be complete in one model and incomplete in the other.
Here's one way to see that (perhaps not the most elegant). Take any reasonable finitely axiomatized theory $T$ in a finite language $L$ capable of sufficient coding to carry out the proof of Gödel's Theorem and not implying the consistency of $ZFC$. (My favorite choice is $I\Sigma_1$, but others might prefer something in the language of set theory.) Fix your favorite encoding of $Con(ZFC)$ in this theory.
Let $T^-$ be some other finitely axiomatized theory on the same language which is complete and so that $T^-\vdash\neg Con(ZFC)$. It need not have any relation to $T$, so it could just interpret all functions as projections onto the first coordinate and all relations as empty, assuming that deduces $\neg Con(ZFC)$. (Note that $\neg Con(ZFC)$ has no meaning in this theory---it's just some arbitrary existential sentence.)
Now consider the sentence $\phi$ which is
$$(Con(ZFC)\rightarrow T)\wedge(\neg Con(ZFC)\rightarrow T^-)$$
(Interpreting $T$ and $T^-$ as the sentences which are the conjunctions of all their axioms.)
In any model of $ZFC$ where $Con(ZFC)$ holds, this is incomplete because $T$ is incomplete. (And also because it's basically a disjunction of two theories and can't tell which case holds.)
In a model $V_1$ of $ZFC$ where $Con(ZFC)$ fails, $T$ can actually prove $\neg Con(ZFC)$, because this is a $\Sigma_1$ formula. Therefore the $T$ branch of the disjunction is self-invalidating: $V_1\vDash T\vdash\neg Con(ZFC)$, so $V_1\vDash \phi\vdash Con(ZFC)\rightarrow\neg Con(ZFC)$, so $V_1\vDash\phi\vdash T^-$. And (the deductive closure of) $T^-$ is complete.
