Every color must appear, and the cyclic conditions red $\to$ blue $\to$ yellow $\to$ red tightly constrain the layout

Count all valid colorings of the fixed (labeled) grid, then reduce by the $8$ rotations and reflections using Burnside’s lemma

First count colorings of the grid with its positions labeled. Checking the $27$ possible rows in succession and rejecting a row as soon as a square whose neighbors are now known lacks its required next color gives the following complete count by the numbers of red, blue, and yellow squares: $$\begin{array}{c|c} (\#R,\#B,\#Y)&\text{valid colorings}\\ \hline (2,4,3)&16\\ (3,2,4)&16\\ (3,3,3)&36\\ (4,3,2)&16 \end{array}$$ Thus the identity symmetry fixes $16+16+36+16=84$ colorings.

For the other symmetries, the fixed-coloring counts are $0$ for each nontrivial rotation, $6$ for each reflection across a horizontal or vertical axis, and $0$ for each diagonal reflection. (For an axis reflection, a direct check of the three palindromic rows gives the $6$ possibilities.) Therefore Burnside’s lemma gives $$\frac{84+2\cdot6}{8}=12.$$

Thus, the correct answer is C.