Question: Looking for an example of a symplectic manifold $(M,\omega)$ that is not symplectomorphic to $(M,-\omega)$.

In particular this means that $M$ must be chiral (i.e. doesn't admit an orientation-reversing diffeomorphism).

For a topological obstruction, I think it would be enough to find $(M,\omega)$ such that $\mathrm{Diff}(M)$ acts trivially on $H^2(M)\neq0$.

A complex projective variety defined by real equations won't work, because the complex conjugation map is antisymplectic.

Answer: 
Let $n \geq 2$ be a natural number and $M$ a torus of dimension $2n$. Then a generic element of $H^2(M, \mathbb R)$ comes from a symplectic form, because we can take a $2$-form invariant under the torus action representing it, and it is a symplectic, and a generic such form is nondegenerate.

Thus it is sufficient to show that for a generic $\omega \in H^2(M,\mathbb R)$, the diffeomorphism group does not send $\omega$ to $-\omega$.

The action of the diffeomorphism group factors through the representation $\wedge^2$ of $GL\_{2n}(\mathbb Z)$. Because $2n>2$, no element acts as $-1$ on this representation: It would have to have the product of any two different eigenvalues be $-1$, but there cannot be $3$ or more numbers with this property.

For each $\sigma \in GL\_{2n}(\mathbb Z)$, the $\omega$ that are sent to $-\omega$ by it form a proper subspace in $H^2(M, \mathbb R)$, and there are countably many such. Choosing $\omega$ outside all these subspaces gets you an example. For instance, a random $\omega$ succeeds with probability $1$.
The following answer was a suggestion of Ivan Smith's. It seems like a very nice argument, although the proof is quite high-tech.

Suppose $X$ is a compact symplectic manifold. By adding a small generic 2-form, we can ensure that the coefficients of $\omega$ with respect to a basis of $H^2(X;\mathbb{Z})$ are linearly independent over $\mathbb{Q}$. Any diffeomorphism that reverses $\omega$ then has to act as $-1$ on $H^2$.

The question now is to find an $X$ such that no element of $\mathrm{Diff}$ acts as $-1$ on $H^2$, and Ivan said that a K3 surface would work. $H^2$ has rank 22, and the signature of the intersection form is $(3, 19)$. Apparently Donaldson theory means that $\mathrm{Diff}$ preserves the orientation of a positive-definite 3-dimensional subspace of $H^2$, so in particular cannot act as $-1$. This is explained in Donaldson-Kronheimer (Corollary 9.1.4), but the basic idea is that a choice of such an orientation allows one to orient some gauge-theoretic moduli space. Reversing the orientation reverses the sign of a corresponding Donaldson invariant, but this invariant is non-zero so it's not equal to itself with sign reversed.