Probabilities in non-stationary states

This is in general true whenever you calculate the projection onto an eigenstate (and not a combination thereof). Let $\left\{|a\rangle\right\}_{a\in A}$ be a set of eigentstates for the Hamiltionian $\hat{H}$, a state at time $t$ can be written as $$ |\psi(t)\rangle = \sum_{a}\hat{U}(t)|a\rangle\langle a |\psi_0\rangle. $$ Its projection onto an eigenstate $|a'\rangle$ is $$ \langle a'| \psi(t)\rangle = \langle a'| \Big(\sum_{a}\hat{U}(t)|a\rangle\langle a |\psi_0\rangle\Big)=\hat{U}(t)_{a' a'}\langle a'|\psi_0\rangle $$ whose norm does not depend on time as long as $\hat{U}(t)$ only picks up a phase factor when acting onto eigenstates. This is because once a state collapses into an eigenstate, it remains there indefinitely.

You just happened to consider an observable (or rather, a pair of observables) that is, in fact, an integral (integrals) of motion of the system. In other words, the probability of measuring any value of $(l,m)$ is in fact not expected to change during time evolution.

This is not true for other observables in general, but it does hold for any time-independent $A$ which commutes with the Hamiltonian. Since both $L^2$ and $L_z$ have this property, both $l$ and $m$ are integrals of motion and your result follows. (They also commute with each other which enables you to use both the measured values simultaneously.)

For a counterexample, you may consider the probability of measuring something that is not an eigenstate of the Hamiltonian, like $|\varphi\rangle := (|1,0,0\rangle + |2,0,0\rangle)/\sqrt2$. I won't try to come up with an observable of which this is an eigenvector – that would only obscure the idea and at the end of the day you only need the eigenvector anyway. If you want, examples of common observables that don't commute with the hydrogen Hamiltonian are any component of position or of momentum, but there the direct calculation is complicated by the fact that these do not have eigenvalues.