Let $n$ be a positive intger. Is the following true? For continuous maps $f: \mathbb S^n \rightarrow \mathbb S^n$ and $g: \mathbb S^n \rightarrow \mathbb R^n$, there exists a point $x \in \mathbb S^n$ such that $g(x) = g(f(x))$.

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
In that same paper, Hopf apparently proved that:
A generalization was obtained by Pannwitz in
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. Update. Still no luck getting Pannwitz's paper. However, I have managed to find the following related paper
(available online here), 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 $g(y) \leq g(f(y))$. The intermediate value theorem then asserts that there is a $z \in \mathbb S^1$ for which $g(z) = g(f(z))$. 


This is not true for $n=2$. Represent $\mathbb S^2$ as the cylinder $\mathbb S^1\times[1,1]$ with two discs $D_+$ and $D_$ attached to the boundary components $\mathbb S^1\times\{1\}$ and $\mathbb S^1\times\{1\}$, resp. Denote by $\mathbb S^2_+$ and $\mathbb S^2_$ the "positive" and "negative" hemispheres: $\mathbb S^2_+=(\mathbb S^1\times[0,1])\cup D_+$ and $\mathbb S^2_$ is the opposite. First we define $g:\mathbb S^2\to\mathbb R^2$. Consider a smooth $\infty$shaped loop $\gamma:\mathbb S^1\to\mathbb R^2$, namely $$ \gamma(t) = (\sin t,\tfrac12\sin2t) $$ (here $\mathbb S^1=\mathbb R/2\pi\mathbb Z$). Note that the velocity of $\gamma$ is separated away from the vertical vector $e_2=(0,1)$, namely $\angle(\dot\gamma(t),e_2)\ge\pi/4$ for all $t\in\mathbb S^1$. Choose $\varepsilon>0$ so small that $\angle (\gamma(t+\varepsilon)\gamma(t),e_2)> \pi/5$ for all $t$. For $x=(t,s)$ from the cylinder, define $$ \begin{cases} g(x) =\gamma(t)+1000s\cdot\overrightarrow{(1,10)} , &\qquad s\ge 0 \cr g(x) =\gamma(t)+1000s\cdot\overrightarrow{(1,10)} , &\qquad s\le 0 . \end{cases} $$ (its image of the cylinder consists of two almost vertical strips above the $\infty$figure). Observe that the image of each boundary component $\mathbb S^1\times\{\pm 1\}$ is separated away from the image of the other half of the cylinder (by distance at least 100). Extend $g$ to $D_+$ and $D_$ so as to fill these boundary components within their neighborhoods of radius 2. Then $g(D_+)\cap g(\mathbb S^2_)=\emptyset$ and $g(D_)\cap g(\mathbb S^2_+)=\emptyset$. Since $\gamma(t+\varepsilon)\gamma(t)$ never forms a small angle with $e_2$, the construction guarantees that $g(t,s)\ne\gamma(t+\varepsilon)$ for all $t\in\mathbb S^1$, $s\in[1,1]$. Now we define $f:\mathbb S^n\to\mathbb S^n$. For $x=(t,s)$ from the cylinder, define $f(x)=(t+\varepsilon,0)$, so the cylinder is projected to its equator and slightly rotated. Extend $f$ to $D_+$ and $D_$ so that $f(D_+)\subset \mathbb S^2_$ and $f(D^)\subset \mathbb S^2_+$. For these $f$ and $g$, we have $g(x)\ne g(f(x))$ for all $x\in\mathbb S^2$. Indeed, if $x=(t,s)\in\mathbb S^1\times[1,1]$, then $f(g(x))=\gamma(t+\varepsilon)\ne g(x)$ as noted above. For $x\in D^+$ this follows from the fact that $g(D^+)\cap g(f(D^+))=\emptyset$, and similarly for $D^$. 


On the other hand, it is known to be true if f is an involution, i.e. f(f(x))=x. This was proved in a paper by C.T. Yang in Annals of Mathematics in 1954. 

