This problem can indeed be solved by Polya counting. First place the eight red beads along a circle, leaving some kind of space between them for the blue beads. Now place two blue beads between each adjacent pair of red beads as well as an empty slot that will be filled later. That leaves $32-2*8 = 16$ blue beads. Now the empty slots may be filled by any number of blue beads as long as there are $16$ blue beads in total. This means that the blue beads going into the eight slots have generating function $ f(z) = \frac{1}{1-z}.$ What we have in fact is that the generating function is $g(z) = 1 + z + z^2 + z^3 + \cdots z^{15} + z^{16}$ but we shall see that we can use $f(z)$ in place of $g(z)$ with no overcounting.
Now the eight slots are being permuted by $C_8$, the cylic group of order $8.$ The cycle index of the cyclic group $C_n$ is $ Z(C_n) = \frac{1}{n} \sum_{d|n} \varphi(d) a_d^{n/d}$ so that $Z(C_8)$ is $ Z(C_8) = \frac{1}{8} \left(a_1^8 + a_2^4 + 2 a_4^2 + 4 a_8 \right).$ It follows that the number of orbits or necklace configurations is given by $ [z^{16}] Z(C_8)_{a_1=f(z); a_2=f(z^2); a_4=f(z^4); a_8=f(z^8)}$ which is $ [z^{16}] \frac{1}{8} \left(\frac{1}{(1-z)^8} + \frac{1}{(1-z^2)^4} + 2 \frac{1}{(1-z^4)^2} + 4 \frac{1}{(1-z^8)} \right).$ Finally recall that $ [z^n] \frac{1}{(1-z)^q} = \binom{n+q-1}{q-1}$ so that the value of the coefficient of $z^{16}$ is $\frac{1}{8} \left( \binom{16+7}{7} +\binom{8+3}{3} +2\binom{4+1}{1} +4\binom{2+0}{0}\right) = 30667.$ In the case where instead of the cyclic group $C_n$ the dihedral group $D_n$ is acting on the slots on the necklace (i.e., the symmetry includes reflections) the cycle index is given by $ Z(D_n) = \frac{1}{2} Z(C_n) + \frac{1}{4} \left( a_1^2 a_2^{(n-2)/2} + a_2^{n/2}\right)$ so that the answer becomes $ \frac{1}{2} 30667 + \frac{1}{4} [z^{16}] \left( \frac{1}{(1-z)^2}\frac{1}{(1-z^2)^3} + \frac{1}{(1-z^2)^4}\right) = 15581.$