A variant of the Stone-Weierstrass theorem? I would like to ask specialists in C*-algebras if the following variant of the Stone-Weierstrass theorem is true.
Suppose $A$ is a C*-algebra and $C$ is its center. Since $C$ is a commutative C*-algebra, there exists a compact space $T$ such that $C$ is isomorphic to the algebra $C(T)$ of continuous functions on $T$. Does this mean that there exists a C*-algebra $B$ such that 
1) $A$ is isomorphic to a closed subalgebra in the algebra $C(T,B)$ of continuous mappings $f:T\to B$ with the pointwise algebraic operations (and the topology of uniform convergence on $T$), and
2) this isomorphism turns $C$ into the algebra of scalar mappings, i.e. the mappings of the form $f(x)=\lambda(x)\cdot 1_B$, where $1_B$ is the identity in $B$, and $\lambda(x)$ $\in$ $\mathbb{C}$ for all $x\in T$.
EDIT 21-03-12: All the C*-algebras here are supposed to be unital, excuse me for not mentioning this from the very beginning!
 A: 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 that the complete regularisation map 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 Blanchard showed that if $A$ is separable and exact and the complete regularisation map is open then such a $B$ can be found (E. Blanchard, Subtriviality of continuous fields of nuclear C*-algebras, J. Reine Angew. Math. 489 (1997), 133-149).
A: The statement of the Dauns-Hofmann theorem is actually too weak to get bundles of $C^*$-algebras. For completeness, let me state it:
Let $A$ be a $C^*$-algebra. For each $P \in Prim(A)$, let $\pi_P \colon A \to A/P$ be the quotient map. Then there is an isomorphism $\phi$ of $C_b(Prim(A))$ onto the center $ZM(A)$ of the multiplier algebra $M(A)$ such that for all $f \in C_b(Prim(A))$ and $a \in A$
$$
\pi_P(\phi(f)a) = f(P)\pi_P(a)
$$
for every $P \in Prim(A)$. Usually one writes $f \cdot a = \phi(f)a$. 
So, the best you could hope for is some kind of sheaf of $C^*$-algebras over the primitive ideal space. Getting local triviality in general is kind of hopeless, I think. A reading recommendation for these matters would be the book "Morita Equivalence and Continuous-Trace $C^*$-algebras" by Raeburn and Williams. 
For continuous trace $C^*$-algebras things are quite different. These are all Morita equivalent (or stably isomorphic) to sections in a bundle of compact operators!
A: EDIT 20-03-12 It seems from the recent answers of Douglas Somerset and Ulrich Pennig that what I claim below is false, and so this answer should be "dis-accepted".

I think (although I admit I don't know the details) that the answer to both questions is yes, by a theorem of Dauns and Hoffman. According to the version quoted in the article
T. Becker, A few remarks on the Dauns-Hofmann theorems for $C^\ast$-algebras.
Archiv der Mathematik 43 (1984) no. 3, 265-269 [Math Review]
$A$ can be realized as the algebra of continuous sections of some kind of continuous $C^\ast$-algebra-bundle with base space $T$.
However, since I am not a specialist, I may have misread or misunderstood.
