This is false for all $n \geq 2$, but true for $n=1$.

A map $g \colon X \to Y$ of topological space is said to be **free** if there is a map $f \colon X \to X$ such that $g(x) \neq g(f(x))$ for all $x\in X$. This definition appears to be due to Hopf, and was given in
> H. Hopf, *Freie Überdeckungen und freie Abbildungen*, Fund. Math. **28** (1937), 33--57.

In that same paper, Hopf apparently proved that:

1. There is a free map $g \colon \mathbb S^n \to \mathbb R^n$ for every $n \geq 3$.
2. There is no free map from $\mathbb S^n$ into $\mathbb R$ if $n \geq 1$.

A generalization was obtained by Pannwitz in
>E. Pannwitz, *Eine freie Abbildung der n-dimensionalen Sphäre in die Ebene.* Math. Nachr. **7** (1952), 183–-185.

According to the MR, she proved that, for every $n \geq 0$, there exists a free map $g \colon \mathbb S^n \to \mathbb R ^2$.

Unfortunately, I'm having trouble accessing the papers, so I can't provide any more details.

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**Update.** Still no luck getting Pannwitz's paper. However, I have managed to find the following related paper
> C. Biasi, D. de Mattos, E. dos Santos, *Applications of the non-standard version of the Borsuk--Ulam theorem*, JP J. Geom. Topol. **9** (2009), no. 3, 273--284.

(available online [here][1]), which mentions the above papers of Hopf and Pannwitz in its Introduction. More interesting however is Theorem 2.1, which I believe can be applied to Sergei's construction to yield Pannwitz's result for $n\geq 2$. This would show that the question in the OP has a negative answer if $n\geq 2$.

It remains to show that the answer is "yes" if $n=1$. (As Harry notes, the answer is also "yes" if $n=0$.) The following is Hopf's argument. Let $g \colon \mathbb S^1 \to \mathbb R$ and $f \colon \mathbb S^1 \to \mathbb S^1$ be given. Since $\mathbb S^1$ is compact, $g$ attains its maximum, resp. minimum, at some $x$, resp. $y$, in $\mathbb S^1$. In particular, $g(x) \geq g(f(x))$ and $g(y) \leq g(f(y))$. Now restrict $g$ to the arc of $\mathbb S^1$ connecting $x$ to $y$. By the intermediate value theorem, there is a $z$ on this arc for which $g(z) = g(f(z))$.


  [1]: http://www.icmc.usp.br/~sma/cadernos/toc8.2/287.pdf