Actually, your claim as stated is wrong even for rational functions.
Indeed, consider the function
$f:z\mapsto i\cdot\frac{z^2-1}{z^2+1}.$
The preimage of the real axis (including $\infty$) under this map is the unit circle $\mathbb{T}$.
However, the map $f:\mathbb{T} \to \mathbb{R} \cup \{\infty\}$ is not injective on the unit circle (it is a 2-1 covering map).
For meromorphic functions, you can go even further: Take the map $f:z\mapsto i\cdot \frac{e^z-1}{e^z+1}.$ (Can you spot a pattern?) The preimage of the real axis here is just the imaginary axis. So once more, this preimage is a simple closed curve when we add in $\infty$, but the map is an infinite-to-one covering map.
However, we can prove the following. I will replace the extended real axis by the unit circle for convenience (in order to get the original statement, just compose with a Möbius transformation as in the examples above).
Theorem. Let $f$ be a nonconstant meromorphic function. Then there is a nontrivial closed curve $\gamma\subset f^{-1}(\mathbb{T})\cup\{\infty\}$ such that $f(\gamma\cap\mathbb{C})$ is a dense subset of $\mathbb{T}$.
Sketch of proof. Let $D$ be the unit disk, and let $V$ be a connected component of $f^{-1}(D)$.
If $V$ is not simply connected, let $\gamma$ be the boundary of one of the complementary components of $V$. It should be easy to see that then $\gamma$ is mapped to $\mathbb{T}$ as a finite-degree covering map.
So suppose that $V$ is simply connected. It is easy to see that the boundary of $V$ is locally connected near every finite point. Since a continuum cannot fail to be locally connected at only one point, it follows that the boundary of $V$ is locally connected. By Carathéodory's theorem, the boundary is the image of a continuous curve $\gamma:\mathbb{T}\to \partial V$. To see that $f(\gamma\cap\mathbb{C})$ is dense in the unit circle, we can simply apply the Gross star theorem. This theorem says that a branch of the inverse of a meromorphic function can be analytically continued along almost every radial ray. This completes the proof.
One can do a closer analysis of the mapping behavior of $f$ on the curve $\gamma$. Of course if $f$ is rational, then we can ensure that $f$ maps $\gamma$ to $\mathbb{T}$ as a finite-degree covering map.