Yes, any such $f$ is constant. In fact, any such $f$ which is bounded below is constant. This can be proven by martingale theory, as can the statement that harmonic functions bounded below are constant ([Liouville's theorem][1]).

Let $X_1,X_2,\ldots$ be a sequence of independent random random variables uniformly distributed on the unit circle, set $S_n=\sum\_{m=1}^nX_m$ and let $\mathcal{F}\_n$ be the sigma-algebra generated by $X_1,X_2,\ldots,X_n$. Then, $S_n$ is a random walk in the plane, and is recurrent. Your condition is equivalent to $\mathbb{E}[f(S_{n+1})\vert\mathcal{F}\_n]=f(S_n)$. That is, $f(S_n)$ is a martingale. It is a standard result that a martingale which is bounded below converges to a limit, with probability one. However, as $S_n$ is recurrent, this only happens if $f$ is constant almost everywhere. By continuity of $f$, it must be constant everywhere.

For the same argument applied to functions $f\colon\mathbb{Z}^2\to\mathbb{R}$, see Byron Schmuland's answer to [this math.SE question][2].


  [1]: http://en.wikipedia.org/wiki/Harmonic_functions#Liouville.27s_theorem
  [2]: http://math.stackexchange.com/q/51926/1321