The answer to this question is YES. I assume you want $A$ and $B$ to be connected. Already in the case of four manifolds two symplectic surfaces can have negative intersection.
To construct an example, we use that if two symplectic surfaces in a $4$-fold intersect positively, you can always smoothen the neighborhood of their intersection to get a new symplectic surface. (Of course this is not always true in the algebraic case).
Example. Take any symplectic 4-manifold $M$ with a symlectic surface $C$, such that $C^2=-1$ (for example $\mathbb CP^2$ blown up once). Blow up $M$ at a point of $C$ and let $E$ be the exceptional curve, while $C'$ be the proper transform of $C$. Notice that $C'^2=-2$ and $C'$ intersect $E$ positively in one point. So we can smoothen the union $C'\cup E$ (this is the full preimage of $C$ under the blow up, and we can not smooth this union algebraically) and get a surface that we call $B$. Finally we have $B\cdot C'$=$C'^2+C'\cdot E=-2+1<0$.
ADDED. Smoothing. In order to smoothen $C'\cup E$ symplectically we can first chose coordinates in a local chart $U$ and adjust a bit the symplectic form so the $C'\cup E$ is given in $U$ by $zw=0$, and the symplectic form is $-i(dz\wedge d\bar z+dw\wedge d\bar w)$. Then surely the curve given by $zw=\varepsilon$ is symplectic in $U$ (since the smoothen curve is complex, while the symplectic form is Kahler). But the whole curve is now discontinuous on the boundary of $U$. To cure this we consider $zw=\varepsilon (F(|z|^2+|w|^2))$ where $F(0)=1$ in $U/2$ and $F$ is smooth with compact support. It is symplectic again, since inside $U\setminus (U/2)$ it is a little perturbation of the union of lines $z=0$ and $w=0$