Here is one way to look at this problem. I will write $I$ for $\bar{R}$.
Firstly, $B/IB = (R/I) \otimes_R B$, that is, $B/IB$ is the image of $B$ under the induction functor (change of rings) $R$-mod $\to R/I$-mod. This functor will be written $\uparrow$. Frobenius reciprocity tells us
\hom_{R/I}( B\uparrow, B'\uparrow) \cong \hom_R (B, B'\uparrow\downarrow) where $\downarrow$ is restriction $R/I$-mod $\to R$-mod. We'll make this an identification, so we consider $\phi$ as an element of the right hand side. The question is then: "is $\phi$ in the image of \hom_R(B,B') \to \hom_R(B,B'\uparrow \downarrow) ?" (the map on homs arises from B' \to B\uparrow \downarrow, the morphism given by the universal property of induction).
This stuff fits into the long exact sequence obtained by applying $\hom_R(B, -)$ to IB' \to B' \to B'\uparrow\downarrow. We get
0\to \hom_R(B, IB') \to \hom_R(B, B') \to \hom_R(B, B'\uparrow\downarrow) \stackrel{\omega}{\to} \operatorname{Ext}^1 _R(B, IB') \to \cdots
$\phi$ being in the image is equivalent to its being in the kernel of the connecting homomorphism $\omega$, which is multiplication by the short exact sequence above. So one answer to your question is: $\phi$ lifts if and only if it is killed by multiplication by the short exact sequence.
Really though, this is just a restatement of the problem. Clearly $B$ being projective is enough to guarantee the lift exists, as is any assumption that makes that ext group \operatorname{Ext}^1 _R(B, IB') vanish. But for arbitrary $\phi$ I don't know general conditions other than the one above. It's sufficient for the induction functor to be full. $R/I$ being flat over $R$ (so induction is exact) certainly isn't enough.