Here is another attempt at proving no such functor exists — I apologize to Chris and to Manny if something like this is already in the papers which they cite.$\newcommand{\Mat}{{\bf M}}\newcommand{\Cplx}{{\bf C}}\newcommand{\Cst}{{\rm C}^*}$

Let $\Mat_2$ denote the algebra of $2\times 2$ complex-valued matrices, and let $D_2$ be the algebra $\Cplx\oplus\Cplx$ regarded as the subalgebra of $\Mat_2$ consisting of diagonal matrices. Let
$$ A = \{ f\in C([0,1], \Mat_2) \mid f(1)\in D_2\}. $$
Since $Z(\Mat_2)=\Cplx$, one shows (with a small appeal to continuity) that
$$ Z(A)= \{ f\cdot I_2 \mid f\in C([0,1])\}\cong C([0,1]).$$

Let $\phi: D_2 \to A$ be the $*$-homomorphic embedding $\phi(d)(t) = d$ for all $t\in [0,1]$ and all $d\in D_2$. Let ${\rm ev}_1 : A\to D_2$ be the homomorpphism "evaluate at $1$".

Now suppose $\mathcal F$ is a functor on the category of $\Cst$-algebras and $*$-homomorphisms, such that ${\mathcal F}(B)=Z(B)$ for every $\Cst$-algebra $B$. Then ${\mathcal F}({\rm ev}_1) \circ {\mathcal F}(\phi)$ must be the identity homomorphism on ${\mathcal F}(D_2)=D_2$, and thus the identity homomorphism on $D_2$ must factor somehow through the algebra $Z(A)\cong C([0,1])$. But this is impossible, since there is no injective algebra homomorphism from $D_2$ into $C([0,1])$. One way to see this last claim is to note that $D_2$ contains two non-trivial projections $e_1$ and $e_2$ which sum to $1$, while the only projections in $C([0,1])$ are $0$ and $1$.