Your claim "for the QHO you can derive that $\langle \hat{x}(t)\rangle =0$ for all $t$" applies only to expectation values of energy eigenstates $|n\rangle$, but *not* for the general case of an arbitrary state vector $$|\psi\rangle = \sum\limits_{n=0}^\infty c_n |n\rangle, \qquad \sum\limits_{n=0}^\infty |c_n|^2 =1.$$ A well known case, where the expectation value of the position operator of the harmonic oscillator performs a "classical" oscillation is the *coherent state* $$   |z\rangle= e^{-|z|^2/2} e^{z a^\dagger} |0\rangle = \sum\limits_{n=0}^\infty\frac{e^{-|z|^2/2} z^n}{\sqrt{n!}} |n\rangle, \qquad z \in \mathbb{C}.    $$