It all comes down to the time derivative of $\hat{H}$. Regardless of being time dependent or independent in Schroedinger picture, one would have to use Heisenberg picture hamiltonian in the equation of motion. Heisenberg picture hamiltonian has vanishing time derivative, so the time evolution of $\hat{V}$ won't have $\dot{\hat{H}}$ in it.

See this thread for details on the time-dependent Hamiltonian and Heisenberg picture.

It all comes down to the time derivative of $\hat{H}$. Regardless of being time dependent or independent in Schroedinger picture, one would have to use Heisenberg picture hamiltonian in the equation of motion. Heisenberg picture hamiltonian has vanishing time derivative, so the time evolution of $\hat{V}$ won't have $\dot{\hat{H}}$ in it.

See this thread for details on the time-dependent Hamiltonian and Heisenberg picture.

Updated answer

In Heisenberg picture equation of motion is:

$$\frac{d}{dt} \hat{A}_H = \frac{i}{\hbar} [ \hat{H}_H, \hat{A}_H ] + \left( \frac{\partial{\hat{A_S}}}{\partial{t}} \right)_H$$

While $\hat{x}$ is a time-independent operator in Schroedinger picture, the velocity may have non-zero partial derivative $\frac{\partial}{\partial{t}}$. The extra term that you get by doing things manually is going to be the same as the last term in the full equation of motion.