Is there a connection $\nabla$ for which this particular non geodesible vector field $X$ satisfy $\nabla_X X=0$? Let $X$  be  the  following  vector  field on $\mathbb{R}^2\setminus \{0\}$
\begin{align}
x' &= x\,(1-x^2-y^2)(x^2+y^2-3) - y\,(2-x^2-y^2)\\
y' &= y\,(1-x^2-y^2)(x^2+y^2-3) + x\,(2-x^2-y^2).
\end{align}
It is  well known that there is  no  a  Riemannian metric on the  punctured plane such that the  orbits  of  the  above  vector  field  would  be  unparametrized  geodesics.  The  proof  is based on existence of  opposite orientation of  two consecutive nested  limit cycles in the phase portrait  of  the vector  field.
Now we  reduce the above  geodesibility requirement to the following  question;(Is this  an obvious reduced question)?

Is there a (torsion free or metric or arbitrary)  connection $\nabla$  on the  punctured plane with $\nabla_X X=0$?

 A: Your vector field can be written in polar coordinates as 
$$X=X^r\partial_r+X^{\varphi}\partial_{\varphi}=r(4r^2-r^4-3)\partial_r+(2-r^2)\partial_{\varphi},$$
which exhibits the rotational symmetry.
The condition $\nabla_X X=0$ implies that
\begin{align}
X^r\partial_rX^r+\Gamma^r_{rr}(X^r)^2+(\Gamma^r_{r\varphi}+\Gamma^r_{\varphi r}) X^rX^{\varphi}+\Gamma^r_{\varphi\varphi}(X^{\varphi})^2=0, \\
X^r\partial_rX^{\varphi}+\Gamma^{\varphi}_{rr}(X^r)^2+(\Gamma^{\varphi}_{r\varphi}+\Gamma^{\varphi}_{\varphi r}) X^rX^{\varphi}+\Gamma^{\varphi}_{\varphi\varphi}(X^{\varphi})^2=0.
\end{align}
For a connection with vanishing torsion, we have $\Gamma^r_{r\varphi}=\Gamma^r_{\varphi r}$ and $\Gamma^{\varphi}_{r\varphi}=\Gamma^{\varphi}_{\varphi r}$, which leads to the simplification
\begin{align}
X^r\partial_rX^r+\Gamma^r_{rr}(X^r)^2+2\Gamma^r_{r\varphi} X^rX^{\varphi}+\Gamma^r_{\varphi\varphi}(X^{\varphi})^2=0, \\
X^r\partial_rX^{\varphi}+\Gamma^{\varphi}_{rr}(X^r)^2+2\Gamma^{\varphi}_{r\varphi} X^rX^{\varphi}+\Gamma^{\varphi}_{\varphi\varphi}(X^{\varphi})^2=0.
\end{align}
Since the vector field has rotational symmetry, I will look for a connection with coefficients that depend only on $r$.
By making a fourth-order polynomial-in-$r$ ansatz for all Christoffel symbols,
$$\Gamma^i_{jk}=\sum_{m=0}^4\Gamma^i_{jkm}r^m,$$
one can find the following solution:
$$\Gamma^r_{rr}=-\frac{r}{2}, \quad \Gamma^r_{r\varphi}=\Gamma^r_{\varphi r}=\frac{1}{4}(3-12r^2+r^4), \quad \Gamma^r_{\varphi\varphi}=0,$$
$$\Gamma^{\varphi}_{rr}=2, \quad \Gamma^{\varphi}_{r\varphi}=\Gamma^{\varphi}_{\varphi r}=r(2-r^2), \quad \Gamma^{\varphi}_{\varphi\varphi}=0.$$
It's worth noting that anything lower than a fourth-order polynomial will not produce any solution.
There does not seem to be any other connection with polynomial coefficients.
