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 $g(x) = g(f(x))$.

OK, so I've been thinking about your problem a little bit, and I'm still a little bit unclear. Why do you call this statement a generalization of BorsukUlam? – Thierry Zell Apr 9 '11 at 17:05

@Thierry: If f is the antipodal map of the sphere, then this is exactly the BorsukUlam Theorem. – Bill Kronholm Apr 9 '11 at 17:12

1Thanks Bill, I guess I'm not very awake this morning! – Thierry Zell Apr 9 '11 at 17:16
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), 3357.
In that same paper, Hopf apparently proved that:
 There is a free map $g \colon \mathbb S^n \to \mathbb R^n$ for every $n \geq 3$.
 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 ndimensionalen 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.
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 nonstandard version of the BorsukUlam theorem, JP J. Geom. Topol. 9 (2009), no. 3, 273284.
(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 $D^$.

Minor note, you seem to have clockwise and counterclockwise backwards. – Harry Altman Apr 10 '11 at 15:10

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.