The examples in Ruan's paper "Symplectic topology on algebraic 3-folds" (JDG 1994) seem to qualify: take any two algebraic surfaces V and W which are homeomorphic but such that V is minimal and W isn't. These are nondiffeomorphic, but VxS2 and WxS2 are diffeomorphic, and Ruan gives lots of examples (starting with V equal to the Barlow surface and W equal to the 8-point blowup of CP2) where the diffeomorphism can be arranged to intertwine the first Chern classes, whence by a theorem of Wall the almost complex structures are isotopic. However, the distinction between the GW invariants between V and W (which holds because V is minimal and W isn't) survives to VxS2 and WxS2, so VxS2 and WxS2 aren't symplectic deformation equivalent.
Gromov--Witten invariants are designed to count the "number" of curves in a space in a deformation invariant way. Since the number of curves can change under deformations, the Gromov--Witten invariants won't have a direct interpretation in terms of actual numbers of curves, even taking automorphisms into account.
Here is an example of how a negative number might come up, though strictly speaking it isn't a Gromov--Witten invariant. Let M be the moduli space of maps from P^1 to a the total space of O(-4) on P^1. Call this space X. Note that I said maps from P^1, not a genus zero curve, so the source curve is rigid. That's why this isn't Gromov--Witten theory. Any such map factors through the zero section (since O(-4) has no nonzero sections), so this space is the same as the space of maps from P^1 to itself. I just want to look at degree one maps, so the moduli space is 3 dimensional.
We could also compute the dimension using deformation theory: the deformations of a map f are classified by $H^0(f^\ast T)$ where T is the tangent bundle of the target. The target in this case is O(-4), not just P^1, and the tangent bundle restricts to O(2) + O(-4) on the zero section. Thus $H^0(f^\ast T)$ is indeed 3-dimensional, as we expected. However, the Euler characteristic of $f^\ast T$ is not 3 but 0, which means that the "expected dimension" is zero.
The meaning of expected dimension is rather vague. Roughly speaking, it is the dimension of the moduli space for a "generic" choice of deformation. The trouble is that such a deformation might not actually exist. Nevertheless, we can still pretend that a generic deformation does exist and, if the expected dimension is zero, compute the number of curves that it "should" have.
What makes this possible is the obstruction bundle E on M. Any deformation of X gives rise to a section of E and the vanishing locus of this section is the collection of curves that can be deformed to first order along with X. Even though a generic deformation might not exist, the obstruction bundle does still exist, and we can make sense of the vanishing locus of a generic section by taking the top Chern class.
In our situation, the (fiber of the) obstruction bundle is $H^1(f^\ast T)$. Since O(2) does not contribute to H^1, the obstruction bundle is $R^1 p_\ast f^\ast O_{P^1}(-4) = R^1 p_\ast O_{P^3 \times P^1}(-4, -4)$ where $p : P^3 \times P^1 \rightarrow P^3$ is the projection. By the projection formula, this is $O(-4)^{\oplus 3}$ and the top Chern class is -64. This is the "Gromov--Witten invariant" of maps from P^1 to $O_{P^1}(-4)$.
Unfortunately, I don't have anything to say about what this -64 means...
Best Answer
There are some very good reasons why the majority of calculations are done for algebraic manifolds. Maybe the most naive reason is as follows: it is harder to solve PDEs than to draw lines through two points in a space. Somehow everyone knows that for two points in $\mathbb CP^n$ there is exactly one line that passes through them, and so a certain GW invariant equals $1$. From the point of view of symplectic geometry this is a completely non-trivial result. Indeed, if you take the standard symplectic structure $w$ on $\mathbb CP^n$ (coming from the Fubini-Study metric), take $J$ that is tamed by $w$ and try to make the calculation, you will first need to know that almost complex curves exist for $J$ locally (which is already non-trivial), then using different compactness arguments will need to homothopy $J$ to the standard complex structure and prove that GW invariant does not change during homothopy, and finally once you get the standard complex structure on $\mathbb CP^n$ perform the above elementary calculation to get $1$ line through any two points. Gromov non-squeezing http://en.wikipedia.org/wiki/Nonsqueezing_theorem is also based on this idea, so all the above is a completely non-trivial task...
A different reason that majority of calculations are done in algebraic case is that up to recently majority of compact symplectic manifolds we constructed from algebraic pieces, by taking Gompf's sum. In this case again on can try first to calculate GW invariants for these pieces and then glue them. Also in small dimensions symplectic manifolds with sufficiently non-trivial GW invariants tend to be algebraic. For example, in dimension 4 it is conjectured that if a GW invariant of rational curves passing trough one point is non-zero then the 4-manifold is a unirulled or rational surface (in particular, it is algebraic).
If you want an explicit calculation for manifolds that are not Kahler, check the following article of McDuff "Hamiltonian S^1 manifolds are uniruled" http://arxiv.org/abs/0706.0675 . She proves non-vanishing of certain GW invariants for manifolds admitting Hamiltonian $S^1$ action and it is not hard to construct such non-Kahler manifolds.
More generally, Tien-Jun Li and Ruan argue in "Symplectic Birational Geometry" http://arxiv.org/abs/0906.3265 that rational GW invariants are responsible for birational geometry of symplectic manifolds.