The answer is *no*.   Suppose that $\alpha_1,\ldots,\alpha_g,\beta_1,\ldots,\beta_g$ were closed $1$-forms on $M$ such that their cohomology classes were a basis of $H^1(M)$ and they satisfied $\alpha_i\wedge\alpha_j = \beta_i\wedge\beta_j = 0$ while $\alpha_i\wedge\beta_j = \delta_{ij}\ \gamma$ where $\gamma$ is a single $2$-form whose cohomology class is nonzero.  Then $\gamma$ cannot vanish identically.  Let $U\subset M$ be the open set on which $\gamma$ is nonzero.  Then none of the $\alpha_i$ or $\beta_i$ can vanish on $U$.  This implies, in particular, that $\alpha_2$ and $\beta_2$ are multiples of $\alpha_1$ on $U$ while $\alpha_1$ and $\beta_1$ are multiples of $\alpha_2$ on $U$.  But this implies that $\alpha_1$ and $\beta_1$ must be linearly dependent on $U$ as well, which implies that $\alpha_1\wedge\beta_1 = 0$.