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.
I upvoted the first answer but I would like to show how to compute the
cycle index $Z(D_6)$ of the dihedral group $D_6$ and apply the
Burnside lemma and the Polya Enumeration Theorem to this problem.
Observe that the OEIS uses the convention of refering to rotational
symmetry as a necklace and as a bracelet when reflectional symmetry is
included, so we are working with a bracelet here.
We need to enumerate and factor the twelve permutations that
contribute to $Z(D_6).$
There is the identity, which contributes
$$a_1^6.$$
The two rotations by a distance of one and five contribute
$$2 a_6.$$
The two rotations by a distance of two or four contribute
$$2 a_3^2.$$
Finally the rotation by a distance of three contributes
$$a_2^3.$$
There are three reflections about an axis passing through opposite
vertices, giving
$$3 a_1^2 a_2^2$$
and three reflections about an axis passing through the midpoints of
opposite edges, giving
$$3 a_2^3.$$
This finally yields the cycle index
$$Z(D_6) = \frac{1}{12}
\left(a_1^6 + 2a_6 + 2a_3^2 + 3a_1^2 a_2^2 + 4a_2^3\right).$$
Do the Burnside calculation first. We have three colors and two
instances of each. The colors must be constant on the cycles. We now
proceed to count these. We get for $a_1^6$ the contribution ${6\choose
2,2,2}.$ There are no candidates for $a_6$ (we do not have six
instances of a color). We do not have three instances either, so zero
for $a_3^2.$ For $a_1^2 a_2^2$ we must choose a pair of colors for the
two-cycles, giving $3\times 2\times {3\choose 2}.$ Finally for $a_2^3$
we get six permutations of the three colors. This yields
$$\frac{1}{12}
\left({6\choose 2,2,2} + 3\times 2\times {3\choose 2}
+ 4\times 6\right)
\\ = \frac{1}{12} (90 + 18 + 24) = 11.$$
On the other hand Polya says that we need the coefficient
$$[A^2 B^2 C^2] Z(D_6)(A+B+C)$$ which is
$$[A^2 B^2 C^2]\frac{1}{12}
\left((A+B+C)^6 + 2(A^6+B^6+C^6) +
2(A^3 + B^3 + C^3)^2 \\+ 3(A+B+C)^2(A^2+B^2+C^2)^2
+ 4(A^2+B^2+C^2)^3\right).$$
Now we may drop the terms that cannot possibly produce multiples of
$A^2 B^2 C^2$ which leaves
$$[A^2 B^2 C^2]\frac{1}{12}
\left((A+B+C)^6
\\ + 3(A+B+C)^2 (A^2+B^2+C^2)^2
+ 4(A^2+B^2+C^2)^3\right).$$
Doing the coefficient extraction then yields
$$\frac{1}{12}
\left({6\choose 2,2,2}
+ 3 \times 3 \times 2 + 4{3\choose 1,1,1}\right) = 11.$$
Here we have used the observation that we must choose a square from
the first term of $(A+B+C)^2 (A^2+B^2+C^2)^2$ and then choose the two
remaining squares from the second factor, which may be done in two
ways.
Best Answer
Well done! I presume you wanted this use of Burnside checked and it is very good.
One could quibble that for an example where there is only one relevant symmetry, applying Burnside seems overly sophisticated. You could just say:
There are $\frac{13!}{6!4!3!}=60060$ arrangements in total, of which $\frac{6!}{3!2!1!}=60$ are symmetrical.
The number of necklaces is therefore $60 + 30000=30060$.
However, as a test of understanding Burnside it was a good easy example to use.