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.