Let $\mu$ be a Borel measure on $X$ that satisfies a zero-one law (i.e. $\mu$ taks only values $0,1$) and has $\mu(X) = 1$. Then $$F: C(X) \to \mathbb{R},\; f \mapsto \int_X f\; d \mu$$ defines a ring homomorphism.
I'm not sure, if there are not even examples for $(X, \mu)$ such that $F$ isn't an evalution (of course in this case $X$ can't be a Q-space). At least, there are such examples for Baire measures.
Added: To give an example, let $\omega_1$ be the first uncountable ordinal and let $X := [0,\omega_1]$ be the set of all ordinals $0 \le \alpha\le\omega_1$ considered as topological space with the order topology. Then $X$ is a compact Hausdorff space. Futhermore, it can be shown that if $B$ is a Borel set, then $$\mu(B) := \begin{cases} 1, \text{ if } B \textstyle \text{ has an unbounded, closed subset of } X \setminus \lbrace \omega_1 \rbrace \newline 0, \textstyle\text{ otherwise}\end{cases}$$ defines a Borel measure on $X$ (cf. Halmos, Measure Theory, Exercise 52.10).
Since $\mu$ is zero on finite sets, it's obviously no dirac measure.
Also note that if $X$ is compact, then, by the Riesz representation theorem, the only non-zero Radon measures with values in $\lbrace 0,1 \rbrace$ are the Dirac measures.