**Update:** I removed the links to the Sage code of the complicated explicit examples, because very much easier examples exist. See below for an example, and this preprint for more details concerning this answer and the computation of explicit examples.

**Answer:** The answer is no. In the following I'll describe how to find two rational functions $f,g\in\mathbb C(X)$, both of odd prime degree $\ell$, such that the following holds:

- $f(g(X))\in\mathbb R(X)$.
- $g(\mathbb R\cup\{\infty\})$ is a smooth closed Jordan curve in the complex plane.

Let $E$ be an elliptic curve, defined over the reals. For $p\in E(\mathbb C)$ we let $\bar p$ be the complex conjugate of $p$. Choose $E$ such that the following holds:

- There is a point $w\in E(\mathbb R)$ with no $y\in E(\mathbb R)$ with $w=2y$.
- There is a point $z\in E(\mathbb C)$ of order $\ell$ with $\bar z\notin\langle z\rangle$. (Such a point always exists.)

Then $C=\langle z\rangle$ is a subgroup of order $\ell$ of $E$, and $E'=E/C$ is an elliptic curve over $\mathbb C$. Let $\phi:E\to E'$ be the associated isogeny, and $\phi':E'\to E$ be the dual isogeny. Then $\phi'\circ\phi:E\to E$ is the multiplication by $\ell$ map.

Let $\beta$ be the automorphism of order $2$ sending $p\in E(\mathbb C)$ to $w-p$. Similarly, define the involutory automorphisms $\beta'$ of $E'$ and $\beta''$ of $E$ by $\beta'(p')=\phi(w)-p'$ and $\beta''(p)=\ell w-p=\phi'(\phi(w))-p$.

Let $\psi$ be the degree $2$ covering map $E\to E/\langle\beta\rangle=P^1(\mathbb C)$, and define likewise $\psi':E'\to E'/\langle\beta'\rangle=P^1(\mathbb C)$ and $\psi'':E\to P^1(\mathbb C)$.

Let $f(X)$ and $g(X)$ be the rational functions defined implicitly by $\psi'\circ\phi=g\circ\psi$ and $\psi''\circ\phi'=f\circ\psi'$. Note that $\psi$, and $\psi''$ are defined over $\mathbb R$, while $\psi'$ is not.

As the multiplication by $\ell$ map $\phi'\circ\phi$ is defined over the reals, we obtain $f(g(X))\in\mathbb R(X)$.

We next claim that $g(X)$ is injective on $\mathbb R$. Suppose there are distinct real $u,v$ with $g(u)=g(v)$. Pick $p,q\in E(\mathbb C)$ with $\psi(p)=u$, $\psi(q)=v$. Then $\psi'(\phi(p))=g(u)=g(v)=\psi'(\phi(q))$, so $\phi(p)=\phi(q)$ or $\phi(p)=\phi(w)-\phi(q)$. Upon possibly replacing $q$ with $w-q$ we may assume $\phi(p)=\phi(q)$, so $p-q\in C$.

Next we study the effect of complex conjugation. As $\psi$ is defined over the reals, and $\psi(p)=u$ is real, we have $\psi(\bar p)=\psi(p)$, so $\bar p=p$ or $\bar p=w-p$. Likewise, $\bar q=q$ or $\bar q=w-q$. Recall that $p-q\in C$, and $\bar C\cap C=\{0\}$ by condition 2. So we can't have $(\bar p,\bar q)=(p,q)$, nor $(\bar p,\bar q)=(w-p,w-q)$.

Thus without loss of generality $\bar p=p$, $\bar q=w-q$. As $p-q$ and $\bar p-\bar q=p-w+q$ have order $\ell$, we see that $2p-w=r$ with $\ell r=0$ and $r\in E(\mathbb R)$. So $w=2(p+\frac{\ell-1}{2}r)$, contrary to condition 1. Furthermore, the function $g(X)$ behaves well at infinity by this geometric interpretation.

An explicit example, with $\omega$ a primitive third root of unity, is
\begin{align*}
f(X) &= \frac{X^3 - 6(\omega + 1)X}{3X^2 + 1}\\
g(X) &= \frac{2X^3 + (\omega + 1)X}{X^2 - \omega}\\
f(g(X)) &= \frac{8X^9 - 24X^5 - 13X^3 - 6X}{12X^8 + 13X^6 + 12X^4
- 1}\in\mathbb Q(X).
\end{align*}

First note that $g(\mathbb R\cup\{\infty\})$ is not contained in a circle, for instance because the points $g(0)=0$, $g(\infty)=\infty$, $g(1/2)=(1+2\omega)/3$, and $g(1)=(5+4\omega)/3$ do not lie on a circle.

Secondly, $g(\mathbb R\cup\{\infty\})$ is a Jordan curve, because $g$ is injective on $\mathbb R$: Suppose that $g(x)=g(x+\delta)$ for real $x,\delta$. A short calculation yields $\delta(8\delta^4 + 14\delta^2 + 49)=0$, so $\delta=0$.

onebranch of the hyperbola $xy=1$ (the one with $x$ and $y$ nonnegative), and this closes to be a (nonsmooth) Jordan analytic curve $\gamma$ on the Riemann sphere. Then the map $f(w) = w^2-2i$ maps $\gamma$ into the real line again. Of course, the composition $f\circ g$ isn't a covering map, but you didn't ask for that. (Credit: This is a slight modification of Noam Elkie's deleted answer.) – Robert Bryant Aug 11 '12 at 14:32