What is the classical counterpart of an eigenstate? Does this question make sense for every system or just some?
If it makes sense, it is a periodic orbit? 
 A: As noted in some of the comments already, the answer to your question depends on what properties of quantum eigenstates you want your classical analogues to have.
If you're thinking of eigenstates of arbitrary observables, quantum eigenstates have the property that the value of that observable is precisely defined. In classical mechanics, individual points in phase space have this property, since the positions and momenta of all particles are precisely defined. One of the interesting things about quantum mechanics is that there are observables (like position and momentum) for which no quantum state can simultaneously be an eigenstate of both.
When a quantum state is not the eigenstate of a particular observable, the value of that observable is not precisely defined. Instead, the quantum state defines a probability distribution over possible values of that observable. You can have the same thing in classical mechanics, if you allow the state of your system to be probability distributions $\rho(\vec{q},\vec{p})$ over points in phase space instead of only single points. This brings us to another possible property of eigenstates: time stationarity.
A quantum mechanical eigenstate of the Hamiltonian will not evolve in time. You might ask yourself what kinds of classical states don't evolve with time. Well, you can see that certain points in phase space will be like this if you set $\dot{q}$ and $\dot{p}$ equal to zero in Hamilton's equations of motion
\begin{align}
\dot{q}&=\frac{\partial H}{\partial p}=\{q,H\}, \\
\dot{p}&=-\frac{\partial H}{\partial q}=\{p,H\}.
\end{align}
(Here the curly braces are Poisson brackets.) These states might not be very interesting (for a 1d simple harmonic oscillator, this state will be where the mass is sitting in the rest position with no velocity). However, if we let our classical states again be distributions over phase space instead of only points, we can get perhaps more interesting states! The evolution of distributions over phase space is governed by the Liouville equation
\begin{align}
\frac{\partial\rho}{\partial t}=-\{\rho,H\}.
\end{align}
Therefore, any distribution $\rho$ over phase space for which $\{\rho,H\}=0$ is an analogue of a stationary state in quantum mechanics. For a 1d simple harmonic oscillator these states include uniform distributions over circles centered at the origin (that is, periodic orbits) in phase space.
The condition $\{\rho,H\}=0$ is very similar to the condition $[\rho,H]=0$ in quantum mechanics, where $[A,B]:=AB-BA$ is the commutator. If $\rho=|\psi\rangle\langle\psi|$ (that is, the quantum state is a pure state and not a probability distribution over pure states), $[\rho,H]=0$ means that $|\psi\rangle$ is an eigenstate of $H$, so we've come full-circle.
As one last thing to leave you with, the equation for describing the time evolution of distributions over quantum states (the von Neumann equation) gives another look at the close relationship between the Poisson bracket in classical mechanics and the commutator in quantum mechanics:
\begin{align}
i\hbar\frac{\partial\rho}{\partial t}=-[\rho,H]
\end{align}
