$S$ = {necklaces of length 14 with 8 blue, 3 green and 3 brown beads}.
Clearly,$|S| = \frac {14!} {8!3!3!} $.
The group of symmetries, $G = D_{14}$. Clearly, $|G| = 28$. And as you have identified G is acting on S.
And by Burnside lemma, required answer is, $$ \#orbits = \frac 1 {|G|}*\sum_{\sigma \in G}fix(\sigma)$$
1) $\sigma = identity $
Then it fixes any element in $S$. Thus, $fix(\sigma) = |S| = \frac {14!} {8!3!3!}$
2) Rotations, $\sigma$ = rotation by $\frac {360} {14} degree$ clockwise = $p^1$
Carefully consider the cyclic structure of the permutation $\sigma$,
$$\begin{pmatrix}
1&2&3&4&5&6&7&8&9&10&11&12&13&14\\
2&3&4&5&6&7&8&9&10&11&12&13&14&1
\end{pmatrix}$$
$$\equiv (1\ 2\ 3\ 4\ 5\ 6\ 7\ 8\ 9\ 10\ 11\ 12\ 13\ 14) \ \ \text{[in cycle notation]}$$
If $\sigma$ fixes $x \in S$, then all vertex of the 14-gon must have the same color which is not the case, hence $fix(\sigma)=0$.
Clearly, $fix(\sigma=p^{13})=0$.
3)$\sigma=p^2$ i.e
$$\begin{pmatrix}
1&2&3&4&5&6&7&8&9&10&11&12&13&14\\
3&4&5&6&7&8&9&10&11&12&13&14&1&2
\end{pmatrix}$$
$$\equiv (1\ 3\ 5\ 7\ 9\ 11\ 13)(2\ 4\ 6\ 8\ 10\ 12\ 14) \ \ \text{[in cycle notation]}$$.
So, $\sigma$ has two cycles of length 7 and if you think carefully, for x to be in $fix(\sigma)$ all vertices in a single cycle will have same bead color. Which is not possible with 8 red, 3 blue and 3 brown beads. So, $fix(\sigma)=0$.
Clearly, $fix(\sigma=p^{12})=0$.
In the same fashion compute the $fix(\sigma)$ for the remaining by looking into the cycle structure and then use the burnside to get the required answer.
You should use Burnside's lemma.
There are $4$ rotations of order $12$. Each of these stabilizes $2$ colorings.
There are $2$ rotations of order $6$. Each of these stabilizes $4$ colorings.
There are $2$ rotations of order $4$. Each of these stabilizes $8$ colorings.
There are $2$ rotations of order $3$. Each of these stabilizes $16$ colorings.
There is $1$ rotation of order $2$. Each of these stabilizes $64$ colorings
There is $1$ rotation of order $1$. It stabilizes the $4096$ colorings.
We now apply Burnside and obtain:
$\frac{4\cdot2+2\cdot4+2\cdot8+2\cdot16+1\cdot64+1\cdot 4096}{12}=352$ necklaces.
Best Answer
We use the Polya Enumeration Theorem (PET). The cycle index $Z(D_6)$ of the dihedral group $D_6$ is given by
$$Z(D_6) = \frac{1}{12} \left(\sum_{d|6} \varphi(d) a_d^{6/d} + 3 a_2^3 + 3 a_1^2 a_2^2\right).$$
This is
$$\frac{1}{12} \left(a_1^6 + a_2^3 + 2 a_3^2 + 2 a_6 + 3 a_2^3 + 3 a_1^2 a_2^2\right) \\ = \frac{1}{12} \left(a_1^6 + 2 a_3^2 + 2 a_6 + 4 a_2^3 + 3 a_1^2 a_2^2\right).$$
We seek
$$[R^2 B^2 Y G] Z(D_6; R+B+Y+G).$$
The only contribution comes from
$$\frac{1}{12} \left(a_1^6 + 3 a_1^2 a_2^2\right).$$
This is because with the Polya substitution the terms $2 a_3^2 + 2 a_6 + 4 a_2^3$ produce powers of three, six, and two, exclusively.
Continuing, we get
$$\frac{1}{12} [R^2 B^2 Y G] (R+B+Y+G)^6 \\ + \frac{1}{4} [R^2 B^2 Y G] (R+B+Y+G)^2 (R^2+B^2+Y^2+G^2)^2.$$
This is
$$\frac{1}{12} {6\choose 2,2,1,1} + \frac{1}{4} \times 2 \times 2 = 16.$$
BTW the convention at the OEIS is to use the term necklace for cyclic symmetries and bracelet for dihedral ones.