I follow Olivier's suggestion and turn my comment on how to realize $\mathbb Q\times \mathbb C$ into an answer.
Fix $r>0$ and consider the ring $\mathbb Q_{r^+}[\![t]\!]$ of power series with coefficients in $\mathbb Q$ that converge on some neighborhood of the closed disc $D(0,r)$ of center 0 and radius $r$ in $\mathbb C$. It is noetherian (see Harbarter, Convergent Arithmetic Power Series). Actually, it is even a PID.
Now consider an evaluation map $f \mapsto f(z)$ with $z \in D(0,r)$. Its image is $\mathbb Q$ if $z=0$, $\mathbb R$ if $z$ is real and $\mathbb C$ otherwise. So we can get $\mathbb Q \times \mathbb C$ as a quotient.
Using the same kind of trick and replacing $\mathbb Q$ by $\mathbb Z$ or localizations of $\mathbb Z$, one should be able to construct a noetherian domain whose quotient is a given finite product of finite fields, finite extensions of $\mathbb Q_p$ or $\mathbb Q$, $\mathbb R$ or $\mathbb C$.
Of course the method has its limits and I have no idea how to realize $\mathbb Q \times \bar{\mathbb Q}$ for instance.