How should I interpret the expectation value $\langle x p\rangle$ in quantum mechanics?

The line from your textbook you're puzzling over is in fact quite a bit easier: in a stationary state nothing changes apart from phases that cancel out in any expectation value, and therefore $\frac{d}{dt}\left[\textrm{anything} \right]=0.$

You're right to point out that $xp$ is not a hermitian operator, but that does not mean that the expectation value $\langle xp \rangle$ is meaningless. Specifically, take the uncertainty relation $xp-px=i\hbar$ and substract $xp$ twice: you get $$xp+px=-i\hbar+2xp,$$ which you can rearrange into $$\langle xp\rangle = \left\langle \frac{xp+px}{2}\right\rangle+\frac{i\hbar}{2}.$$ The expectation value is now of the hemitian operator $\frac{1}{2}(xp+px)$, and you can see that your original expectation value has a trivial imaginary part.

If it's a physical interpretation for this quantity you're after, try this question.

Here is how you interpret: the expectation value in a state $|\Psi\rangle$ of products of Hermitian operators like $x$ and $p$ (corresponding to observables) quantifies how correlated (quantum mechanically entangled) the two observables are in that state.

Therefore, the statement $\frac{d}{dt}\langle xp\rangle=0$ in English reads: "the extent to which the two observables 'position' and 'momentum' are entangled correlated does not change with time". Now since stationary states are independent of time (up to a phase), it should be pretty clear that the correlation should not change with time (i.e. it is fixed).