Is it true that if the average of a continuous function $f:\mathbb{R}^2\rightarrow[0,1]$ over a unit circle centered around $(x,y)$ is $f(x,y)$ for all $(x,y)\in\mathbb{R}^2$, then $f$ is necessarily constant?

Yes, any such $f$ is constant. In fact, if we relax the condition so that $f$ is only required to be bounded below, but not above, then it is still true that $f$ is constant. This can be proven by martingale theory, as can the statement that harmonic functions bounded below are constant (Liouville's theorem). 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 sigmaalgebra 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. In general, for a continuous function $f\colon\mathbb{R}^2\to\mathbb{C}$, if $f(x,y)$ is the average of $f$ on the unit circle centered at $(x,y)$ then it does not follow that $f$ is harmonic. So, we cannot prove the result directly by applying Liouville's theorem. As an example (based on the comments by Gerald Edgar and by me), consider $f(x,y)=\exp(ax)$. The average of $f$ on the unit circle centered at $(x,y)$ is $$ \frac{1}{2\pi}\int_0^{2\pi}f(x+\cos t,y+\sin t)\,dt=\frac{1}{2\pi}f(x,y)\int_0^{2\pi}e^{a\cos t}\,dt=f(x,y)I_0(a). $$ Here, $I_0(a)$ is the modified Bessel function of the first kind. Whenever $I_0(a)=1$ then $f$ satisfies the required property. This is true for $a=0$, in which case $f$ is constant, but there are also nonzero solutions such as $a\approx1.88044+6.94751i$. In that case $f$ satisfies the required property but is not harmonic. 


These functions are called harmonic functions. One the simplest examples is $f(x,y)=xy$. More generally, the real part of any holomorphic function $\mathbb C\to \mathbb C$ is a harmonic function $\mathbb C\to \mathbb R$. Added later: As mentioned by Gerald, harmonic functions are characterized by the property that $$\int_0^1f(z+re^{2\pi\theta})d\theta=f(z),\qquad \forall r\ge 0,\quad \forall z\in \mathbb C.$$ I don't know whether that property for $r=1$ implies that property for all $r\ge 0$. Partial answer to the edited question: If you require the function to be bounded, then I think that yes, that should force it to be constant. Liouville's theorem states that any bounded holomorphic function $\mathbb C\to \mathbb C$ is constant. There is also a version of Liouville's theorem for harmonic functions, so yes: the function is constant. Gap in the argument: 

