We can treat this system classically because it is one of those nice situations in which the quantum-mechanical treatment produces the same results!
I will begin with part 2 of your question about plane waves. The use of this Ansatz is the first clue that you are actually treating the situation quantum-mechanically, but ending up with a result that exactly matches the classical result.
Basically what we are doing when we represent a quantum-mechanical wavefunction $\psi({r})$ as a plane wave $\psi({r}) = e^{-i\mathbf{p} \cdot \mathbf{r}/\hbar}$ is treating a single component of its Fourier transform,
\begin{equation}
\psi{(\mathbf{p})}=\int{d^3\mathbf{p} \: e^{-i\mathbf{p} \cdot \mathbf{r}/\hbar} \: \psi{(\mathbf{r})}}.
\end{equation}
This is mathematically far easier to do than treating the entire wavepacket, and since Quantum Mechanical operators are linear, the behaviour of the plane wave under its operators tells us the behaviour of the entire wavefunction when we sum up all the plane waves under the integral.
Furthermore, we can always do this, since square-integrability, that is,
\begin{equation}
\int^{\infty}_{-\infty}{d^3\mathbf{r} \: |\psi(\mathbf{r})|^2}< \infty
\end{equation}
is a sufficient requirement for the Fourier transform of a function to exist, and obviously this must be true in Quantum Mechanics, since the LHS of the inequality represents the probability of finding our 'particle' anywhere in space, and so must equal $1$.
This is the underlying reason why the use of plane-waves is ubiquitous in Quantum Mechanics. It's easy, and it works.
In your case, you have written $v(t) = v_0 e^{-iEt+i\mathbf{p} \cdot \mathbf{x}}$ where $v_0$ is the classical velocity (the velocity of the midpoint of the quantum-mechanical wavepacket), obtained by taking the FT of $p_x \psi(\mathbf{r}) = \partial_x \psi(\mathbf{r})$, integrating by parts under the Fourier integral, and neglecting boundary terms, to get
\begin{equation}
\int{d^3\mathbf{p} \: p_x e^{-i\mathbf{p} \cdot \mathbf{r}/\hbar} \: \psi{(\mathbf{r})}}.
\end{equation}
Which we can then interpret classically as $mv_0$.
Now for the cyclotron resonance in Quantum Mechanics.
To begin with, I will take the simple case of a free electron of momentum $\mathbf{p}=(p_x,p_y,p_z)$ in a magnetic field along the z-axis, $\mathbf{B}=(0,0,B)$. We can treat this quantum-mechanically by setting the gauge of the electromagnetic vector potential to the Landau gauge
\begin{equation}
\mathbf{A} = (-By,0,0)
\end{equation}
Where $\nabla \times \mathbf{A}=\mathbf{B}$. The Schroedinger equation in this gauge can then be written down by taking the canonical momentum $\mathbf{p} \rightarrow \mathbf{p}-q\mathbf{A}$ in this gauge so that
\begin{equation}
H \psi (\mathbf{r}) = \frac{\hbar^2}{2m}[(p_x+qBy)^2+p_y^2+p_z^2] \psi(\mathbf{r}) = E\psi(\mathbf{r})
\end{equation}
This Hamiltonian clearly commutes with both $p_x$ and $p_z$ which are therefore conserved quantities and so share a set of eigenstates. Taking our plane-wave Ansatz, the $x$ and $z$ parts can be commuted through $H$, so remain as $e^{i(p_x x + p_z z)/\hbar}$, while the $y$ part, which we may call $\chi(y)$, now obeys the new equation
\begin{equation}
\Big[\frac{p_y^2}{2m} + \frac{p_z^2}{2m} + \frac{1}{2} m\omega_c^2 \Big(y+\frac{p_x}{qB}\Big)^2 \Big] \chi(y) = E\chi(y),
\end{equation}
where $\omega_c = |q|B/m$, the classical cyclotron frequency.
We can extend this argument to crossed electric and magnetic fields by introducing an electric potential $\phi = -Ey$ so that we have an electric field $\mathbf{E} = (0,E,0)$ in the $y$-direction, and the Hamiltonian becomes
\begin{equation}
H = \frac{\hbar^2}{2m}[(p_x+qBy)^2+p_y^2+p_z^2] + qEy
\end{equation}
Where we can still use the same Ansatz $e^{i(p_x x + p_z z)/\hbar}\chi(y)$ since H still commutes with $p_x$ and $p_z$. With this substitution we find that we can rearrange back to the cyclotron form we had before, if we switch to a frame in which $p_x \rightarrow p_x + mE/B$. That is:
\begin{equation}
\Big[\frac{p_y^2}{2m} + \frac{p_z^2}{2m} + \frac{1}{2} m\omega_c^2 \Big(y+\frac{p_x-mE/B}{qB}\Big)^2 \Big] \chi(y) = E\chi(y),
\end{equation}
Therefore, this predicts an additional 'drift velocity' in the $x$-direction of
\begin{equation}
p_x = \frac{mE}{B},
\end{equation}
which we can interpret for the classical case of $\mathbf{p}=m\mathbf{v}$ as an actual 'drift' velocity (again the 'velocity' of the centre of the quantum-mechanical wavepacket),
\begin{equation}
v_x = \frac{E}{B}.
\end{equation}
This corresponds again, exactly to what you would get by using the classical Lorentz force equation, $\mathbf{F} = q(\mathbf{E}+\mathbf{v} \times \mathbf{B})$.
We can then introduce dielectrics with given relaxation times $\tau$ as in your question.
Note about the presence of $\hbar$:
Notice here that when we arrive at the equation of motion from the Schroedinger equation, the $\hbar$ factors cancel, making our result independent of $\hbar$ so that we don't need to take the usual classical limit $\hbar \rightarrow 0$. This is not true in all situations, and is what makes this situation special.