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.
You are correct, you want to use Burnside's lemma:
$$\mbox{# of Orbits}=\frac{1}{|G|}\sum_{g\in G}|fix(g)|.$$
You have an action of $G=D_{2m}$ on the set $X$ of $n$-colored necklaces with $m$ beads, and two such necklaces are the same if they are in the same orbit under this action. Therefore, counting the number of orbits counts the number of necklaces.
Let's set up some notation. We can regard a necklace as a regular $n$-gon with vertices numbered $\{1,\ldots,m\}$. Then, $G$ act on the vertices in the natural way. This corresponds to an embedding $G\hookrightarrow S_m$. Identifying an element $g\in G$ with its image in $S_m$, let $\psi(g)$ denote the number of disjoint cycles in the cycle decomposition of $g$.
Let $C$ be a set of $n$ colors (i.e. $|C|=n$). Then $G$ acts on $X=C^m$ by $$g.(c_1,\ldots,c_m)=(c_{g.1},\ldots,c_{g.m}).$$
Lemma: For the action of $G$ on $X$, we have $|fix(g)|=n^{\psi(g)}$.
proof: Since, for all $k$, $g^k.(c_1,\ldots,c_m)=(c_{g^k.1},\ldots,c_{g^k.m})$ we see that $(c_1,\ldots,c_m)$ is in $fix(g)$ if, and only if $c_i=c_{g^k.i}$ for all $i$ and $k$. This just says that $(c_1,\ldots,c_m)\in fix(g)$ if, and only if the coloring is constant on the cycles of $g$. There are $n$ choices to color each cycle, leading to $n^{\psi(g)}$ total colorings that are fixed by $g$.
Example: Suppose $m=4$ and $n=3$. The element of $D_8$ and their cycles types are chosen as follows:
$$
\begin{array}{llll}
1=(1)(2)(3)(4)&r=(1234)
&r^2=(13)(24)&r^3=(4321)\\
j=(24)(1)(3)&rj=(12)(34)&r^2j=(13)(2)(4)&r^3j=(14)(23)
\end{array}
$$
The corresponding orders of $fix(g)$ are given by
$$
\begin{array}{llll}
n^4&n&n^2&n\\
n^3&n^2&n^3&n^2
\end{array}
$$
Therefore, the number of necklaces with $n$ colors is
$$\frac{1}{8}(n^4+2n^3+3n^2+2n).$$
Substituting $n=3$ yields $21$ distinct necklaces.
Best Answer
The problem of calculating the number $p_n$ of necklaces when reflections are included among the admissible symmetries is just as easy, as the equivalence classes are still of the same size $2n$, giving $$p_n = \frac{n!}{2n} = \frac{1}{2} (n-1)!.$$
To compute this with Polya counting we need the cycle index $Z(D_n)$ of the dihedral group $D_n$, which contains reflections and rotations. It is given by $$Z(D_n) = \frac{1}{2} Z(C_n) + \begin{cases} \frac{1}{2} a_1 a_2^{(n-1)/2}, & n \mbox{ odd, } \\ \frac{1}{4} \left( a_1^2 a_2^{(n-2)/2} + a_2^{n/2} \right), & n \mbox{ even.} \end{cases}.$$ There is more information here.
The cycle index computation confirms the result of the basic analysis:
with(numtheory); with(group): with(combinat): pet_cycleind_dihedral := proc(n) local d, s, t; s := 0; for d in divisors(n) do s := s + phi(d)*a[d]^(n/d); od; if type(n, odd) then t := n*a[1]*a[2]^((n-1)/2); else t := n/2*(a[1]^2*a[2]^((n-2)/2)+a[2]^(n/2)); fi; (s+t)/2/n; end; pet_varinto_cind := proc(poly, ind) local subs1, subs2, polyvars, indvars, v, pot, res; res := ind; polyvars := indets(poly); indvars := indets(ind); for v in indvars do pot := op(1, v); subs1 := [seq(polyvars[k]=polyvars[k]^pot, k=1..nops(polyvars))]; subs2 := [v=subs(subs1, poly)]; res := subs(subs2, res); od; res; end; v := proc(n) option remember; local p, k, gf; p := add(cat(q, k), k=1..n); gf := expand(pet_varinto_cind(p, pet_cycleind_dihedral(n))); for k to n do gf := coeff(gf, cat(q, k), 1); od; gf; end;The code in action looks like this:
Remark July 4 2018. Let me observe once more that with all beads a distinct color all orbits are the same size namely $2n$, giving the answer $(n-1)!/2.$ Polya counting is not needed here because Burnside says that if we have $n$ different colors and a color is to be constant on the cycles we need a permutation that factors into at least $n$ cycles. There is only one of these namely the identity which has $n$ fixed points. The number of ways of assigning the $n$ colors to these fixed points is $n!$ and we once more obtain $n!/2/n.$ On the other hand the cycle index becomes useful when we seek to classify bracelets according to the distribution of colors where colors may be repeated. For example with three colors red green and blue and a bracelet of six beads we get
$$Z(D_6)(R+B+G) = {B}^{6}+{B}^{5}G+{B}^{5}R+3\,{B}^{4}{G}^{2}+3\,{B}^{4}GR \\ +3\,{B}^{4}{R}^{2}+3\,{B}^{3}{G}^{3}+6\,{B}^{3}{G}^{2}R \\ +6\,{B}^{3}G{R}^{2}+3\,{B}^{3}{R}^{3}+3\,{B}^{2}{G}^{4} \\ +6\,{B}^{2}{G}^{3}R+11\,{B}^{2}{G}^{2}{R}^{2}+6\,{B}^{2}G{R}^{3} \\ +3\,{B}^{2}{R}^{4}+B{G}^{5}+3\,B{G}^{4}R+6\,B{G}^{3}{R}^{2} \\ +6\,B{G}^{2}{R}^{3}+3\,BG{R}^{4}+B{R}^{5}+{G}^{6}+{G}^{5}R \\ +3\,{G}^{4}{R}^{2}+3\,{G}^{3}{R}^{3}+3\,{G}^{2}{R}^{4} \\ +G{R}^{5}+{R}^{6}.$$
This is where the Maple code may be deployed, using the command
The reader may want to verify some of these using pen and paper. For example the term $3B^4G^2$ represents the possibilities using four blue beads and two green ones which are: green beads adjacent, green beads with one blue bead and three blue beads separating them and green beads with two blue beads separating them.