Is the pre-Lie operad a nonsymetric operad Can a pre-Lie algebra be modeled as a nonsymmetric operad?
Looking at the relation $(x \circ y) \circ z - x\circ (y \circ z) = (x \circ z) \circ y - x \circ (z \circ y ) $
it doesn't look like it could be non symmetric since the variables changes order.
But I found this paper https://hal.archives-ouvertes.fr/hal-00337068/document so it does seems to be both a nonsymmetric operad as well as a symmetric one. Is an algebra over the non-symetric one also a pre-Lie algebra? How does this work?
 A: You are asking two subtly different questions in the title of your question and in the body of your question so I'm going to answer them both.

Is the pre-Lie operad a nonsymetric operad

The preLie operad is usually defined as a symmetric operad. But whenever you have a symmetric operad, you can just forget about the symmetric group actions, and you get a non-symmetric operad. This is what Bergeron and Livernet do in their article: they take the usual preLie (symmetric) operad, forget the symmetric group action, and study it as a non-symmetric operad.
If it helps, you can think about it this way: imagine you have a ring $(R, +, \times)$, you forget its product and you just study the abelian group $(R, +)$. It's a bit similar.

Can a pre-Lie algebra be modeled as a nonsymmetric operad?

I do not believe this is possible. More precisely, I do not believe that there exists a non-symmetric operad $P$ such that the category of ("non-symmetric") $P$-algebras is equivalent to the category of preLie algebras.
For example, if you take the (symmetric) preLie operad and forget about the symmetric group actions to get a non-symmetric operad, then the algebras over this non-symmetric operad (defined as a space $V$ equipped with a map $\mathsf{preLie} \to \mathrm{End}_V$ which is compatible with the operadic composition maps but which is not necessarily equivariant with respect to the action of the symmetric group) are not necessarily preLie algebras. Indeed, by the theorem of Bergeron and Livernet, such a thing is just a space equipped with a certain number of multilinear maps that do not satisfy any relations (because the non-symmetric preLie operad is free). A preLie algebra will be an algebra over this non-symmetric operad, because if you have a morphism of symmetric operad $\mathsf{preLie} \to \mathrm{End}_V$ you can just forget about the symmetric group actions and it will still be a morphism of non-symmetric operads, but the converse is not true.
Maybe it helps to look at a simpler example. Take the (symmetric) operad encoding commutative algebras, $\mathsf{Com}$. As you may know one has $\mathsf{Com}(n) = \Bbbk$ for all $n$, assuming we work in vector spaces, and the symmetric group action is trivial on all $\mathsf{Com}(n)$. Now if you forget about the symmetric group actions, you get a non-symmetric operad $P$ given by $P(n) = \Bbbk$ for all $n$. What are (non-symmetric) algebras over this (non-symmetric) operads? Well as you probably know they are just associative algebras! So forgetting about the symmetric group actions has changed the category of algebras. This is exactly what happens for the preLie operad too: it's a symmetric operad whose algebras are preLie algebras, but if you forget about the symmetric group actions you get a non-symmetric operad with completely different algebras.
