Let $(\Omega, \mathcal{B}(\Omega), \mu)$ be a probability space, $B$ be a separable Banach space. Let $\phi : \Omega \to B$ be a bounded (say by $C$) measurable function, which is Bochner integrable for $\mu$.
Let now $\nu \ll \mu$ be a probability measure, is $\phi$ Bochner integrable for $\nu$?
Answer to the new version of the question (assuming $\nu(B)=1$): This is true, since any bounded function is integrable with respect to a finite measure. The absolute continuity condition is entirely irrevelant.
Answer to the old version of the question (no assumption on $\nu(B)$): No, this isn't even true for the finite-dimensional case. Just take $\Omega=\Bbb R$, $\phi\equiv 1$, $\mu=\mathcal{N}(0,1)$ and $\nu$ to be Lebesgue's measure.