The  answer to this  question is  **yes.** There  is  a complex  polynomial  vector  field on $\mathbb{C}^2$ which  has a complex  limit  cycle  which  does  not  intersect the  real plane $im(z)=im(w)=0$.

Consider the  differential  equation $$\begin{cases}z'=w+(z^2+w^2-4i)\\ w'=-z+(z^2+w^2-4i) \end{cases}$$

The regular  leaf $L: z^2+w^2=4i$ of  this  singular  foliation does  not intersect  the  real part of  $\mathbb{C}^2$. This  leaf, which is  topologically a  cylinder,  has  a  non trivial holonomy. In  fact  we  have  more: there  is  a  closed curve on this  leaf whose  corresponding holonomy  map  is  a hyperbolic  map:it is  not  tangent to the identity map. Here is  the  argument:

The hyperbolicity, hence non triviality, of the  holonomy of  this  leaf is  a   consequence of  [Theorem 3.2  Page 333 of  the  paper:  First  Variation of  Holomorphic  forms  and  some  applications](https://projecteuclid.org/euclid.hokmj/1351257968).

**Elaboration:** The  foliation is defined by $$\omega= (w+(z^2+w^2-4i))dw-(-z+(z^2+w^2-4i))dz=0$$

To  apply the   theorem 3.2  in the above  paper we  find  a  $1-$ form $\alpha$ which  satisfies $d\omega=\omega \wedge \alpha$. 

Represent  the  above  $1$- form  $\omega$ in the  form  $\omega=Pdw-Qdz$. Then for $$\alpha=(P_z+Q_w)/(P^2+Q^2)(Pdz+Qdw)$$ we  have  $d\omega=\omega \wedge \alpha$.  Note  that $P^2+Q^2$  does  not  vanish  on $L$

To  compute  this  integral  we  parametrize the  cylinder $L$ with  
$$ \phi(t)= \begin{cases} z(t)=t+i/t\\w(t)=t/i-1/t  \end{cases}$$ where  $\phi$ is  defined on $t\in \mathbb{C} \setminus \{0\}.$

A  very simple  computation shows  that $\int_{S^1} \phi^* \alpha$ is  non zero  **since we  obtain  a  pole of  order 1  at the  origin.** In  fact  the  later  integral is  $\int_{S^1} 2(z(t)+w(t))dt$. So we  have a non degenerate  pole  that is  a  Pole  of  order  1.  so the  integral  does not  vanish. so  the  leafd $L$  is hyperbolic. $\square$