Primitive ideals of minimal tensor product

I expect that this answer is satisfactory, although it isn't a complete answer. This is really the best result one can hope for.

For a $C^\ast$-algebra $A$ let $Prime(A)$ be the prime ideal space (defined exactly as the primitive ideal space, but with prime (two-sided closed) ideals instead). It is well-known that $Prime(A) = Prim(A)$ when $A$ is separable or when $A$ is Type I.

The following result is due to Proposition 2.16 and 2.17 in [Blanchard, Etienne; Kirchberg, Eberhard Non-simple purely infinite C∗-algebras: the Hausdorff case. J. Funct. Anal. 207 (2004), no. 2, 461–513.] and builds on Kirchberg's deep work on exact $C^\ast$-algebras:

Theorem: If $A$ and $B$ are $C^\ast$-algebra and at least one of them is exact, then $Prime(A\otimes_{\textrm{min}} B)$ is homeomorphic to $Prime(A) \times Prime(B)$.

There is a canonical map $Prime(A) \times Prime(B) \to Prime(A\otimes_{\mathrm{min}} B)$ given by $(I,J) \mapsto I\otimes_{\mathrm{min}} B + A \otimes_{\mathrm{min}} J$, and this is the homeomorphism in the theorem above.

If $I$ and $J$ are primitive ideals in $A$ and $B$ respectively, then $I\otimes_{\mathrm{min}} B + A\otimes_{\mathrm{min}} J$ is also primitive. Hence we get the following corollary for primitive ideal spaces instead of prime ideal spaces.

Corollary: Let $A$ and $B$ be $C^\ast$-algebras for which all prime ideals are primitive (e.g. separable/Type I/simple), and such that at least one of $A$ and $B$ is exact. Then $Prim(A\otimes_{\textrm{min}} B)$ is homeomorphic to $Prim(A) \times Prim(B)$.

The results fail in general (even when $A$ is simple, non-exact, and $B=\mathcal B(\ell^2(\mathbb N))$), so you can't really hope for anything better.

As commutative $C^\ast$-algebras are exact, the above theorem is applicable if one of the $C^\ast$-algebras is commutative.

Note that one gets $Prim(A\otimes_{\mathrm{min}} B) \cong Prim(A) \times Prim(B)$ (with primitive ideal spaces), if, for instance, $A$ is commutative (hence Type I) and $B$ is separable.