If the system is not interacting with some "environment" and the dynamics is purely hamiltonian, superposition states will not relax to stationary states, even if they are "close" to some such stationary state. Instead they evolve on "orbits".
Say, for instance, that $\psi_n$, $\psi_m$ are eigenstates (stationary states) of hamiltonian $H$,
$$
H\psi_n = E_n \psi_n\\
H\psi_m = E_m \psi_m
$$
A state $\psi$ "close" to $\psi_n$, of the form
$$
\psi = A(\psi_n + \epsilon \psi_m)
$$
where $A$ is a normalization constant and $|\epsilon|<< 1$ is a small "overlap" parameter, will evolve in time according to
$$
\psi(t) = A\left(e^{-\frac{i}{\hbar}E_nt}\psi_n + \epsilon e^{-\frac{i}{\hbar}E_mt}\psi_m \right) =\\
= A e^{-\frac{i}{\hbar}E_nt}\left(\psi_n + \epsilon e^{-\frac{i}{\hbar}(E_m-E_n)t}\psi_m\right) =\\
= A(t)\left(\psi_n + \epsilon(t)\psi_m \right)
$$
where $|A(t)| = |A|$ and $|\epsilon(t)| = \epsilon$. Obviously $\psi(t)$ is of the same superposition type and is still "in the neighborhood" of $\psi_n$, but will never decay to $\psi_n$.
Even in the presence of a perturbation $V$, $\psi$ will not "decay" to a stationary state unless it happens to be itself a stationary state of the perturbed hamiltonian $H+V$. Otherwise, it will evolve on an "orbit" determined now by $H+V$ instead of $H$.
What the adiabatic theorem points out is that if a perturbation $V$ is turned on infinitely slowly in time, then a stationary state of $H$, say $\psi_n$, will evolve into a stationary state of $H+V$, say $\Psi_n$. This adiabatic switching-on is equivalent not so much to a continuous fast relaxation from "non-equilibrium" to "equilibrium", but rather to a continuous adjusting of an "orbit". The difference may seem subtle when dealing with large many-body systems, but looking at it this way keeps things consistent regardless of system size.
The "relaxation to equilibrium" picture is more adequate when the system interacts with an environment and its local dynamics is no longer hamiltonian, but dissipative.
