Three thoughts on this. The first is that $A$ probably has to be assumed unital to guarantee that $T$ is compact.
Assuming then that $A$ is unital, each point $t\in T$ corresponds to a maximal ideal $M_t$ of $C$ which generates a closed two-sided ideal $G_t$ in $A$. The ideals ${G_t: t\in T}$ are called the Glimm ideals (after James Glimm who used them in the case when $A$ is a von Neumann algebra). For each element $a\in A$, the mapping $t\mapsto \Vert a+G_t\Vert$ is upper semi-continuous but not in general continuous. Indeed these norm funcions are all continuous if and only if the 'complete regularisation' map from the primitive ideal space of $A$ with the hull kernel topology to $T$ is an open map (R-Y Lee, 1970s). The second thought, therefore, is that a necessary condition for the answer to the question to be yes is probably that the complete regularisation map needs to should be open.
Even when the complete regularisation map is open, I expect that one can find examples where the answer to the question is no, although no such example comes to mind just now [in fact, see E. Kirchberg, S.Wassermann, Operations on continuous bundles of C*-algebras, Math. Ann. 303 (1995), 677-697]. The third thought, however, is that in the 1990s Etienne Blanchard showed that if $A$ is (I think) separable and exact and the complete regularisation map is open then such a $B$ can be found (I think that it is $B(H)$). I have probably garbled Blanchard's result but it would be worth looking up E. Blanchard, Subtriviality of continuous fields of nuclear C*-algebras, J. Reine Angew. Math. 489 (I think it has 'sub-trivial' in the title)1997), 133-149).
