You can use the principle of inclusion-exclusion. Let $s(i_1,i_2,\ldots,i_k)$ denote the number of ways to partition $2N$ vertices into $N$ unordered pairs if $\{i_1,i_1+1\}$, $\{i_2,i_2+1\}$, ..., $\{i_k,i_k+1\}$ (with arithmetic performed mod $2N$) are required to be among the pairs. Observe that this quantity is zero if $i_k=i_j+1 \pmod{2N}$ for any $j$, $k$, since $\{i_j,i_j+1\}$ and $\{i_j+1,i_j+2\}$ cannot both be pairs. Then the number of partitions in which none of the pairs are edges of the cycle graph is s()-\sum_i s(i)+\sum_{i
Now if $s(i_1,i_2,\ldots,i_k)$ is nonzero then it equals $\dfrac{(2N-2k)!}{2^{N-k}(N-k)!}=(2N-2k-1)!!$ since the pairings are fixed for $2k$ of the vertices, which leaves $2N-2k$ vertices to pair up. The number of ways of choosing $\{i_1,i_2,\ldots,i_k\}$ so that $i_k\ne i_j+1\pmod{2N}$ holds for all $j$, $k$, is $\binom{2N-k}{k}+\binom{2N-k-1}{k-1}$. (This is the same as the number of words that can be formed from $k$ Ps and $2N-2k$ Ss or from $k-1$ Ps and $2N-2k$ Ss. Here P stands for "pair" and S stands for "singleton". The word PPSPS, for example, corresponds to the $\{i_1,i_2,i_3\}=\{0,2,5\}$. The $i_j$ are the positions of the Ps, with Ps thought of as occupying two sites and Ss as occupying one. The first binomial coefficient counts the $\{i_1,i_2,\ldots,i_k\}$ in which none of the $i_j$ equals $2N-1$; the second counts the $\{i_1,i_2,\ldots,i_k\}$ in which one of the $i_j$ equals $2N-1$.)
Putting this all together, the number of partitions in which none of the pairs are edges of the cycle graph is $ \sum_{k=0}^N(-1)^k(2N-2k-1)!!\left[\binom{2N-k}{k}+\binom{2N-k-1}{k-1}\right]. $ For $N=1,2,3,4,\ldots$ we get 0, 1, 4, 31, 293, 3326, 44189, 673471, 11588884, 222304897. This is A003436 in the Online Encyclopedia of Integer Sequences, which contains many useful references about this sequence.