The reason for your conflicting results has to do with the subtleties of hermiticity on finite intervals.
Look carefully at the formal steps in the derivation of the Ehrenfest theorem:
$$
\frac{d}{dt} \langle \psi(t) | x |\psi(t)\rangle = \langle \frac{d\psi}{dt} | x |\psi\rangle + \langle \psi | x |\frac{d\psi}{dt}\rangle = \frac{i}{\hbar} \left[ \langle H\psi | x|\psi\rangle - \langle \psi | x H |\psi\rangle\right]
$$
Usually at this point one makes use of the hermiticity of $H$ and proceeds to rearrange the first term so as to obtain
$$
\frac{i}{\hbar} \left[ \langle H\psi | x|\psi\rangle - \langle \psi | x H |\psi\rangle\right] = \frac{i}{\hbar} \langle \psi | [H, x] |\psi\rangle = …= \frac{1}{m} \langle \psi | p |\psi\rangle
$$
And it is no problem to check that $H$ is indeed hermitic.
Here's the catch however: in this particular case it is hermitic on the space of functions periodic on $[0,L]$. But let us look closer at
$$
\langle H\psi \;|\; x\;|\;\psi\rangle \equiv \langle H\psi \;|\; x\psi\rangle = \langle x\psi \;| \;H\psi\rangle^*
$$
It is the matrix element of $H$ between one periodic function, $\psi$, and one that is no longer periodic, $x\psi$, which falls outside its proper domain. For this combo $H$ is no longer hermitic, as we can easily check by direct calculation:
$$
\langle H\psi \;|\; x\psi\rangle \sim - \int_0^L{dx\; \frac{d^2\psi^*}{dx^2} x \psi} = - \int_0^L{dx\; \frac{d}{dx}\left(\frac{d\psi^*}{dx} x \psi\right)} + \int_0^L{dx\; \frac{d\psi^*}{dx} \frac{d}{dx}\left(x\psi\right)} = \\
= - \frac{d\psi^*}{dx} x \psi \big|_0^L + \int_0^L{dx\; \frac{d}{dx} \left[\psi^* \frac{d}{dx}(x\psi)\right]} - \int_0^L{dx\; \psi^*\frac{d^2}{dx^2}\left( x \psi\right) } =
$$
$$
= \left[ \psi^* x \frac{d\psi}{dx} - \frac{d\psi^*}{dx} x \psi \right] \big|_0^L - \int_0^L{dx\; \psi^*\frac{d^2}{dx^2}\left( x \psi\right) }
$$
where the first term on the last line was simplified based on the periodicity of $\psi$. But in what is left of it, the presence of $x$ spoils the periodicity and it no longer vanishes as one would naively hope.
Compare to the case when $\psi$ vanishes on the boundary, as for a particle in an infinite box: the boundary term disappears, or in other words, $x\psi$ is still in the domain of $H$ and the theorem is fine.
