First question
If we start with the Liouville equation, define entropy as $\mathcal{H}$, and take its time derivative, we obtain your last equation: $$ \frac{\mathrm{d} \mathcal{H}}{\mathrm{d}t} = \frac{\mathrm{d}}{\mathrm{d}t}\int -\rho\log{\rho} \, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q} = \int\mathrm{d}\vec{p}\,\mathrm{d}\vec{q} \, (-\log\rho - 1) \frac{\mathrm{d}\rho}{\mathrm{d}t} = 0 $$ Since $\mathrm d \rho/\mathrm d t$ is zero at all times, so is $\mathrm d\mathcal H/\mathrm d t$, which proves that the entropy of distribution $\rho$ remains constant at all times. This is true not only for the equilibrium distribution but for any arbitrary distribution $\rho$. It means that the Liouville equation preserves the entropy of the the initial distribution, i.e., it is capable of describing isentropic processes only. We cannot use the Liouville equation to describe the approach to equilibrium, i.e., the transition from a state of lower entropy to a state where the entropy is higher.
The condition $\mathrm d\mathcal H/\mathrm d t$ does not maximize entropy. Entropy maximization is to be done not with respect to time but with respect to $\rho$. The derivative that must be set to zero is $$ \frac{\mathrm d\mathcal H}{\mathrm d \rho} = 0 $$ subject to $$ \int \rho\, \mathrm{d}\vec{p}\,\mathrm{d}\vec{q} = 1 $$ along with any other constraints that define the macroscopic state.
Second question
Two comments here. First, the variation of $\rho$ with $q_a$ is not $\mathrm d\rho/\mathrm d q_a$ but $\partial \rho/\partial q_a$. Second, to calculate $\mathrm d\rho/\mathrm d q_a$ start with the differential $\mathrm d\rho$: $$ \mathrm d\rho = \frac{\partial\rho}{\partial t}\mathrm d t + \frac{\partial\rho}{\partial q_a}\mathrm d q_a + \cdots $$ then divide by $\mathrm d q_a$: $$ \frac{\mathrm d\rho}{\mathrm d q_a} = \frac{\partial\rho}{\partial t}\frac{1}{\dot q_a} + \frac{\partial\rho}{\partial q_a} $$