Suppose you have a Hamiltonian $H$ with no explicit time dependence. We can treat the Hamiltonian like any other observable operator: Start from the definition of the operator in the <PERSON> picture, then apply the unitary transformation $U = e^{-iHt/\hbar}$. Normally this gives $A(t) = e^{iHt/\hbar} A e^{-iHt/\hbar}$ where $A$ was the time-independent operator in the <PERSON> picture and $A(t)$ the time-dependent operator in the <PERSON> picture. But in this case $A$ is just $H$, which commutes with $e^{-iHt/\hbar}$, so we can cancel out the time dependent factors and end up with $H(t) = H$.
So we see the time dependence cancels out, and in fact this will still hold true even when you express $H$ in terms of other operators, which will themselves have time dependence in the <PERSON> picture. 526
For example, consider the harmonic oscillator Hamiltonian $H = \frac{p^2}{2m} + \frac{1}{2}m\omega^2x^2$. In the <PERSON> picture one gets $x(t) = x(0) \cos(\omega t) + \frac{p(0)}{m \omega} \sin(\omega t)$ and $p(t) = p(0) \cos(\omega t) - m \omega x(0) \sin(\omega t)$. So we have $$\frac{p^2}{2m} = \frac{p(0)^2}{2m}\cos^2(\omega t) + \frac{m \omega^2 x(0)^2}{2}\sin^2(\omega t) - \omega x(0) p(0) \cos(\omega t) \sin(\omega t)$$ and $$\frac{1}{2}m\omega^2x^2 = \frac{m \omega^2 x(0)^2}{2}\cos^2(\omega t) + \frac{p(0)^2}{2m}\sin^2(\omega t) + \omega x(0) p(0) \cos(\omega t) \sin(\omega t)$$ and thus $$H = \frac{p(0)^2}{2m}\cos^2(\omega t) + \frac{m \omega^2 x(0)^2}{2}\sin^2(\omega t) - \omega x(0) p(0) \cos(\omega t) \sin(\omega t) \+ \frac{m \omega^2 x(0)^2}{2}\cos^2(\omega t) + \frac{p(0)^2}{2m}\sin^2(\omega t) + \omega x(0) p(0) \cos(\omega t) \sin(\omega t) \ = \frac{p(0)^2}{2m} + \frac{m \omega^2 x(0)^2}{2}$$
So again the time dependence cancels out. Although that's just one example, my point is that you can still have a time-independent Hamiltonian even if it's expressed in terms of operators that are themselves time-dependent. I think that addresses your concern. 955