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.
No, nor for a dodecahedron. A core problem in both these cases is that regular pentagons can't be embedded in any $\mathbb{Z}^n$.
Proof: Suppose we had such a pentagon $ABCDE$. Then the points $0$, $B-A$, $(B-A)+(E-D)$, $(B-A)+(E-D)+(C-B)$, $(B-A)+(E-D)+(C-B)+(A-E)$ also form a pentagon with integer coordinates and a shorter side length. (We drew a five-pointed star whose diagonals were the side lengths of our first pentagon.) So by infinite descent, there's no smallest pentagon and thus no pentagon at all.
The same argument generalizes to all regular polygons with a number of sides other than $3,4,$ or $6$ and to all other non-dense lattices.
Best Answer
While the cycle index of the rotation group of a Platonic solid can be connected to the cycle index of solid itself, it cannot fully determine it: The cube and octahedron are duals and therefore have the same rotation group ($S_4$), but the two solids have different cycle index polynomials. Likewise the dodecahedron and icosahedron which share the rotation group $A_5$.
\begin{align*} A_5: \quad \frac{1}{60}\left(z_1^5 + 15z_1z_2^2 + 20z_1^2z_3 + 24z_5\right) \\ \text{Dodecahedron:} \quad \frac{1}{60}\left(z_1^{12} + 15z_2^6 + 20z_3^4 + 24z_1^2z_5^2\right) \\ \text{Icosahedron:} \quad \frac{1}{60}\left(z_1^{20} + 15z_2^{10} + 20z_1^2z_3^6 + 24z_5^4\right) \end{align*}
To derive the icosahedron cycle index from that of $A_5$, you need to think about what five things are being permuted (five tetrahedra inside the icosahedron) and how permuting them affects the faces of the solid.
For good measure, here are the analogous cycle index polynomials for the other dual pair of Platonic solids.
\begin{align*} S_4: \quad \frac{1}{24}\left(z_1^4 + 3z_2^2 + 6z_1^2z_2 + 6z_4 + 8z_1z_3\right) \\ \text{Cube:} \quad \frac{1}{24}\left(z_1^6 + 3z_1^2z_2^2 + 6z_2^3 + 6z_1^2z_4 + 8z_3^2\right) \\ \text{Octahedron:} \quad \frac{1}{24}\left(z_1^8 + 9z_2^4 + 6z_4^2 + 8z_1^2z_3^2\right) \end{align*}
In the cube, the four things being permuted are diagonals connecting antipodal vertices. In the octahedron, it's the four segments connecting the midpoints of opposite faces. Permuting each of those four things will change the faces of the respective solids differently, thus the different cycle indexes.
At least the cycle indexes for $A_4$ and the tetrahedron are the same ($\frac1{12}[z_1^4+8z_1z_3+3z_2^2]$).
For a reference, see Biggs Discrete Mathematics (Oxford, 2002) ch. 27, "Symmetry and Counting." The cycle indexes for the Platonic solids are summarized in Table 27.3.2 on p. 397.