Here is a small comment to Seva's answer, which allows to get smaller bounds for some fields.

Assume that we managed to find functions $f,g$ so that $\sum f(a)W(a)=0$ if $\deg W\leqslant \alpha$, $\sum g(a) W(a)=0$ if $\deg W\leqslant \beta$. Then we have 
$$
\sum_{a\in A,b\in B} f(a)g(b)Q(a,b)=0
$$ 
for any polynomial $Q$ of degree at most $\alpha+\beta+1$. Indeed, if $Q$ is monomial, then our sum factorizes and one of factors is 0. 

Apply this for a polynomial $Q(x,y)=P(x-y)$, where $\deg P\leqslant \alpha+\beta+1$ and $P(a,b)=0$ for $a,b\in A$, $a\ne b$. We get
$$
0=\sum_{a\in A,b\in B} f(a)g(b)P(a-b)=P(0)\cdot \sum f(a)g(a).
$$
Now if $\sum f(a)g(a)\ne 0$, we conclude that $P(0)=0$. Assume for simplicity that $\alpha=\beta$. Then functions $f,g$ satisfying our condition are chosen in the same linear space of dimension at least $|A|-\binom{n+\alpha}\alpha$ (or $|A|-\sum_{i\leqslant \alpha} \binom{n}i$, if we consider only multilinear polynomials). If $P(0,0)\ne 0$, then this space is an isotropic subspace of $\mathbb{F}^{A}$ (any two vectors must be orthogonal.) Maximal dimension of isotropic subspace is well known subject and it always does not exceed $|A|/2$ be obvious reasons (such a space is contained in its own orthogonal complement). For real field there is no isotropic subspace even of dimension 1. 

See the answer by Robin Chapman to my [old question here][1] for references on isotropic subspaces.


  [1]: http://mathoverflow.net/questions/38304/maximal-number-of-mutually-orthogonal-vectors