All possible arrangements of the beads in the circle are 9!. If we start from a specific bead to which we assign number 1 (starting point), then the number of possible arrangements of the 3 different colors is $\binom{9}{3}\binom{6}{3}\binom{3}{3} = \frac{9!}{3!3!3!}$
It seems to me that it is impossible for ALL beads to have neighboring beads of the same color - am I missing something?
Not all red beads have a neighbor of other color if and only if the red beads are placed on consecutive spots.
Let $R$ denote the event that the red beads are placed on consecutive spots.
Let $B$ denote the event that the blue beads are placed on consecutive spots.
Let $G$ denote the event that the green beads are placed on consecutive spots.
Then to be found is $1-\mathsf P(R\cup B\cup G)$ and with inclusion/exclusion and symmetry we find that this equals:
$$\begin{aligned}1-3\mathsf{P}(R)+3\mathsf{P}(R\cap B)-\mathsf{P}(R\cap B\cap G) & =1-\mathsf{P}\left(R\right)\left[3-3\mathsf{P}\left(B\mid R\right)+\mathsf{P}(B\cap G\mid R)\right]\\ & =1-\frac{9}{\binom{9}{3}}\left[3-3\frac{4}{\binom{6}{3}}+\frac{2}{\binom{6}{3}}\right]\\ & =\frac{41}{56}\\ & \approx0,732142857 \end{aligned} $$
Observe that there are $\binom93$ distinct triples of spots and by $9$ of them the spots are consecutive.
This explains $\mathsf P(R)=9/\binom93$.
Under condition of event $R$ there are $6$ consecutive spots left so there are $\binom63$ triples in total of which $4$ are consecutive and $2$ of which are such that the events $G$ and $B$ will both occur ifthey are possessed by e.g. blue beads.