In fact, the opposite is true. For $X$ a smooth proper variety over $\mathbb C$ and $\omega$ a nonzero symplectic form on $X$. Then there is no nonempty Zariski open set on which $\omega$ is the pullback (under any map) of the standard symplectic form. The same should work for etale open sets, and even for smooth morphisms.

The reason is that the standard symplectic form is exact (i.e. is the derivative of a 1-form), so its cohomology class vanishes - we can just use ordinary de Rham cohomology for this. Thus its pullback under any map has vanishing cohomology class.

But it is not possible for any nonzero holomorphic $2$-form on a smooth projective variety $X$ to have nonzero cohomology class when restricted to any open set $U$. If it did, then the associated class in $H^2(X, \mathbb C)$ would lie in the image of the natural map from the $H^2$ of $X$ supported in $X \setminus U$. But this cohomology group is generated by the classes of the $\dim X-1$-dimensional irreducible components of $X \setminus U$, which are sent to their divisor classes. So the associated class in $H^2(X, \mathbb C)$ would have to be a linear combination of divisor classes.

But divisor classes are sent by Hodge theory to $(1,1)$-forms, while holomorphic $2$-forms are $(0,2)$-forms, so they can never be equal.

In fact, the opposite is true. For $X$ a smooth proper variety over $\mathbb C$ and $\omega$ a nonzero symplectic form on $X$. Then there is no nonempty Zariski open set on which $\omega$ is the pullback (under any map) of the standard symplectic form. The same should work for etale open sets, and even for smooth morphisms.

The reason is that the standard symplectic form is exact (i.e. is the derivative of a 1-form), so its cohomology class vanishes - we can just use ordinary de Rham cohomology for this. Thus its pullback under any map has vanishing cohomology class.

But it is not possible for a nonzero holomorphic $2$-form on a smooth projective variety $X$ to have zero cohomology class when restricted to a nonempty open set $U$. If it did, then the associated class in $H^2(X, \mathbb C)$ would lie in the image of the natural map from the $H^2$ of $X$ supported in $X \setminus U$. But this cohomology group is generated by the classes of the $\dim X-1$-dimensional irreducible components of $X \setminus U$, which are sent to their divisor classes. So the associated class in $H^2(X, \mathbb C)$ would have to be a linear combination of divisor classes.

But divisor classes are sent by Hodge theory to $(1,1)$-forms, while holomorphic $2$-forms are $(0,2)$-forms, so they can never be equal.