Answer
If $t=R+G+B$ then the number of arrangement with no adjacent identical colour balls is
$$\sum_{k=0}^{t}(-1)^{k}\sum_{k_R+k_B+k_G=k}\binom{R-1}{k_R}\binom{B-1}{k_B}\binom{G-1}{k_G}\frac{(t-k)!}{(R-k_R)!(B-k_B)!(G-k_G)!}$$
Explanation
For set of red balls $\{r_1,r_2\ldots,r_R\}$, blue balls $\{b_1,b_2\ldots,b_B\}$, and green balls $\{g_1,g_2\ldots,g_G\}$ call an "adjacency" of an arrangement of the balls a single occurrence of equal colour adjacent balls. Then, if $A_k$ is the sum of intersection of sets of arrangements with at least $k$ adjacencies, then by inclusion-exclusion the required count is
$$\sum_{k=0}^{t}(-1)^k|A_k|\, .\tag{1}$$
So it simply remains to show that $|A_k|$ is the inner summation in the answer.
Consider that, for any set $S$ of identical coloured balls, there are
$$\binom{|S|-1}{k_{|S|}}\frac{|S|!}{(|S|-k_{|S|})!}\tag{2}$$
ways to place elements into $|S|-k_{|S|}$ blocks of unordered elements with $k_{|S|}$ adjacencies: Adjacent elements within blocks contribute an adjacency but those adjacent elements in neigbouring blocks do not e.g. $(r_1,r_2,r_3)(r_4,r_5)$ has adjacency $r_1,r_2$ but not $r_3,r_4$ and is different from $(r_2,r_1,r_3)(r_4,r_5)$ but the same as $(r_4,r_5)(r_1,r_2,r_3)$. The elements adjacent inside the blocks contribute exactly $1$ adjacency, so our examples here each have $3$ adjacencies.
To see $(2)$, notice there are $\binom{|S|-1}{k_{|S|}}$ ways to put $|S|$ identical balls in $|S|-k_{|S|}$ block with each block containing at least $1$ ball. Then there are $|S|!$ ways of labelling those balls. But we must divide out the repeats, which is simply the number of ways of arranging $|S|-k_{|S|}$ blocks i.e. divide by $(|S|-k_{|S|})!$, hence $(2)$.
For $k=k_R+k_B+k_G$ adjacencies there are $\binom{R-1}{k_R}\frac{R!}{(R-k_R)!}$ ways of forming $R-k_R$ red blocks with $k_R$ adjacencies, $\binom{B-1}{k_B}\frac{B!}{(B-k_B)!}$ ways of forming $B-k_B$ blue blocks with $k_B$ adjacencies and $\binom{G-1}{k_G}\frac{G!}{(G-k_G)!}$ ways of forming $G-k_G$ green blocks with $k_G$ adjacencies. There are then $(R+G+B-(k_R+k_B+k_G))!=(t-k)!$ ways of arranging all $t-k$ blocks, giving
$$\binom{R-1}{k_R}\binom{B-1}{k_B}\binom{G-1}{k_G}\frac{R!B!G!(t-k)!}{(R-k_R)!(B-k_B)!(G-k_G)!}$$
total arrangements with at least $k$ adjacencies and at least $k_R$ red contributions, at least $k_B$ blue contributions and at least $k_G$ green contributions.
For each $k$ this must be summed over possible values of $k_R,k_B$ and $k_G$. Thus the total number of arrangements of those sets with no equal colour adjacencies is, by $(1)$
$$\sum_{k=0}^{t}(-1)^{k}\sum_{k_R+k_B+k_G=k}\binom{R-1}{k_R}\binom{B-1}{k_B}\binom{G-1}{k_G}\frac{R!B!G!(t-k)!}{(R-k_R)!(B-k_B)!(G-k_G)!}\, ,$$
but if we want to treat red balls as identical, blue balls as identical and green balls as identical then we must divide out the arrangements of those $R!B!G!$ giving our answer at the top.
So, let's use the example of $R=B=G=3$ then we have:
$$\begin{multline}\frac{9!}{3!^3}-3\binom{2}{1}\binom{2}{0}^2\frac{8!}{2!3!^2}+\left(3\binom{2}{2}\binom{2}{0}^2\frac{7!}{1!3!^2}+3\binom{2}{1}^2\binom{2}{0}\frac{7!}{3!2!^2}\right)\\-\left(6\binom{2}{2}\binom{2}{1}\binom{2}{0}\frac{6!}{1!2!3!}+\binom{2}{1}^3\frac{6!}{2!^3}\right)+\left(3\binom{2}{1}^2\binom{2}{2}\frac{5!}{2!^21!}+3\binom{2}{2}^2\binom{2}{0}\frac{5!}{1!^23!}\right)\\-3\binom{2}{2}^2\binom{2}{1}\frac{4!}{1!^22!}+\binom{2}{2}^3\frac{3!}{1!^3}=174\end{multline}$$
Note that this method is really just the Laguerre Polynomial method in disguise but I thought it might be instructive to present it this way as it sheds some light on the underlying combinatorics.
