Complexity of finding 2 vertex-disjoint $(|V|/2)$-cycles in cubic graphs?
I assume, as stated in the comments, that the problem is to find a 2-factor with two cycles of length $|V|/2$ each in a cubic graph.
This problem is NP-complete:

1) HC in cubic graphs is NP-complete.

2) The following problem HC' is also NP-complete: Given a cubic graph $G$ and an edge $e$
in $G$, is there a Hamiltonian cycle through $e$. We can reduce HC to it: Given a graph $G$, replace one arbitrary vertex (left hand side) by a triangle with another vertex in it (right hand side):



Call this new graph $G'$.
$G'$ is Hamiltonian iff $G$ is. Furthermore, if $G'$ is Hamiltonian, then for any red edge, there is a Hamiltonian cycle that contains this red edge. (Without the vertex in the middle, every Hamiltonian cycle contains two green edges and any green edge can then be replaced by two red ones.) The reduction from HC to HC' maps a cubic graph $G$ to $G'$ together with any red ege.

3) Finally, we reduce HC' to the problem of the question. Given an instance
$(G,e)$, we create two copies of $G$, subdivide $e$ in each copy and connected the two vertices that subdivide the two copies by a new edge $f$. This edge $f$ is a bridge in the new graph. If $G$ has a Hamiltonian cycle through $e$, then the new graph has a 2-factor consisting of two cycles of the same size.
Conversely, if the new graph has such a 2-factor, then it cannot use $f$. Therefore, it splits into two Hamiltonian cycles of the two copies. Since in these copies, the vertex subdividing $e$ has degree two, we get a Hamiltonian cycle of $G$ that contains $e$.