This question has a definite answer, which is to use the Polya Enumeration Theorem, which is more powerful than Burnside and includes it as a special case. It will produce a multivariate generating function for all distributions of colors, from which you may extract the desired coefficient for a given configuration. You compute the cycle index of the face permutation group $G$ of the polyhedron and evaluate it using the substitution
$$a_k = X_1^k + X_2^k + \cdots + X_n^k$$
if you have $n$ colors. Then you extract the coefficient for your particular configuration, which is written like this:
$$[X_1^2 X_2^2 \cdots X_n^2] Z(G)(X_1+X_2+\cdots+X_n).$$
Here are two simple examples where we restrict ourselves to rigid motions (no reflections). Suppose we have a tetrahedron whose faces we want to color. The cycle index of $G$ for this case contains the identity, which contributes $$a_1^4.$$ Then there are rotations about axes passing through the middle of a face and the opposing vertex (four such axes), which fix that face and contribute $$4\times 2\times a_1 a_3.$$ Finally there are three flips by 180 degrees about an axis passing through the midpoints of two vertex-disjoint edges (three such axes), which contribute $$3\times a_2^2.$$ This gives the cycle index
$$Z(G) = \frac{1}{12} (a_1^4 + 8 a_1 a_3 + 3 a_2^2).$$
We can recover Burnside from this via
$$Z(G)(X_1+X_2+\cdots+X_n)_{X_1=X_2=\cdots=X_n=1}$$
which gives the formula
$$\frac{1}{12} (n^4 + 8 n^2 + 3 n^2) = \frac{1}{12} n^4 + \frac{11}{12} n^2.$$
for colorings of the tetrahedron with at most $n$ colors.
This sequence is documented as OEIS A006008.
Now to answer your question, suppose we have three colors $R$, $G$ and $B$ and we want the number of colorings where two colors are used twice. This is where the substituted cycle index comes into play. Making the substitution as described above we have
$$Z(G)(R+B+G)\\ =
1/12\, \left( R+G+B \right) ^{4}+2/3\, \left( R+G+B \right) \left( {R}^{3
}+{G}^{3}+{B}^{3} \right) +1/4\, \left( {R}^{2}+{G}^{2}+{B}^{2} \right) ^{
2}$$
which expands to
$${B}^{4}+{B}^{3}G+{B}^{3}R+{B}^{2}{G}^{2}+{B}^{2}GR+{B}^{2}{R}^{2}+B{G}^{3}
+B{G}^{2}R\\+BG{R}^{2}+B{R}^{3}+{G}^{4}+{G}^{3}R+{G}^{2}{R}^{2}+G{R}^{3}+{R}
^{4}.$$
Extracting coefficients we see that
$$[R^2 G^2]Z(G)(R+G+B)+[R^2 B^2]Z(G)(R+G+B)+[G^2 B^2]Z(G)(R+G+B) = 3,$$
so there are three colorings up to rotation of the faces of the tetrahedron with three available colors using two colors twice.
As a second example consider painting the faces of a cube with three colors. We need another cycle index call it $Z(H)$, the one of the face permutation group of the cube, which is actually a fairly standard computation. There is the identity, which contributes $$a_1^6.$$ There are rotations about an axis passing through opposite faces (three such axes) which fix those faces, giving $$3\times (2 a_1^2 a_4 + a_1^2 a_2^2).$$ There are rotations about axes passing through opposite vertices (four such axes), giving $$4\times 2 \times a_3^2.$$ Finally there are rotations about axes passing through the midpoints of pairs of opposite edges (six such pairs), which contribute $$6\times a_2^3.$$
This gives the cycle index
$$Z(H) = \frac{1}{24}
(a_1^6 + 6 a_1^2 a_4 + 3 a_1^2 a_2^2 + 8 a_3^2 + 6 a_2^3).$$
We recover Burnside from this as above, getting
$$\frac{1}{24} (n^6 + 6 n^3 + 3 n^4 + 8 n^2 + 6 n^3)
= \frac{1}{24} (n^6 + 3 n^4 + 12 n^3 + 8 n^2)
\\= \frac{1}{24} n^6 + \frac{1}{8} n^4 + \frac{1}{2} n^3 + \frac{1}{3} n^2.$$
This is sequence is OEIS A047780.
Now to get the colorings using each color twice we compute the substituted cycle index
$$Z(H)(R+G+B)=
1/24\, \left( R+G+B \right) ^{6}+1/4\, \left( R+G+B \right) ^{2} \left( {R
}^{4}+{G}^{4}+{B}^{4} \right) \\+1/8\, \left( R+G+B \right) ^{2} \left( {R}^
{2}+{G}^{2}+{B}^{2} \right) ^{2}+1/3\, \left( {R}^{3}+{G}^{3}+{B}^{3}
\right) ^{2}+1/4\, \left( {R}^{2}+{G}^{2}+{B}^{2} \right) ^{3}$$
which expands to
$${B}^{6}+{B}^{5}G+{B}^{5}R+2\,{B}^{4}{G}^{2}+2\,{B}^{4}GR+2\,{B}^{4}{R}^{2}
+2\,{B}^{3}{G}^{3}\\+3\,{B}^{3}{G}^{2}R+3\,{B}^{3}G{R}^{2}+2\,{B}^{3}{R}^{3}
+2\,{B}^{2}{G}^{4}+3\,{B}^{2}{G}^{3}R+6\,{B}^{2}{G}^{2}{R}^{2}\\+3\,{B}^{2}G
{R}^{3}+2\,{B}^{2}{R}^{4}+B{G}^{5}+2\,B{G}^{4}R+3\,B{G}^{3}{R}^{2}+3\,B{G}
^{2}{R}^{3}+2\,BG{R}^{4}\\+B{R}^{5}+{G}^{6}+{G}^{5}R+2\,{G}^{4}{R}^{2}+2\,{G
}^{3}{R}^{3}+2\,{G}^{2}{R}^{4}+G{R}^{5}+{R}^{6}.$$
Extracting coefficients we now see that
$$[R^2 G^2 B^2]Z(H)(R+G+B) = 6,$$
so there are six colorings using each of the three colors exactly twice.
The following MSE link points to a chain of similar computations.
