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First of all, it is not true that there are only two types of nondegenerate $3$-forms in $\Lambda^3(V^\ast)$ when $\dim_\mathbb{R}V=6$. There are actually $3$ such nondegenerate orbits. Here is the complete list of $\mathrm{GL}(V)$-orbit types in this space: Let $e^1,\ldots, e^6$ be a basis of $V$. Then the following $3$-forms are inequivalent under $\mathrm{GL}(V)$ and every element of $\Lambda^3(V^\ast)$ is $\mathbb{GL}(V)$-equivalent to exactly one of these:

1. $\phi_1 = e^1\wedge e^2\wedge e^3 + e^4\wedge e^5\wedge e^6$
2. $\phi_2 = e^1\wedge e^3\wedge e^5 - e^1\wedge e^4\wedge e^6-e^2\wedge e^3\wedge e^6- e^2\wedge e^4\wedge e^5$
3. $\phi_3 = e^1\wedge e^5\wedge e^6 + e^2\wedge e^6\wedge e^4+e^3\wedge e^4\wedge e^5$
4. $\phi_4 = e^1\wedge e^2\wedge e^5 + e^3\wedge e^4\wedge e^5$
5. $\phi_5 = e^1\wedge e^2\wedge e^3$
6. $\phi_6 = 0$

The first three types are nondegenerate, and the first two types have open $\mathrm{GL}(V)$orbits, while the orbit of the third type is a hypersurface in $\Lambda^3(V^\ast)$.

The dimension of the annihilators are $0$, $0$, $0$, $1$, $3$, and $6$, respectively. The stabilizers are

1. $\mathrm{Stab}^+(\phi_1) = \mathrm{SL}(3,\mathbb{R})\times\mathrm{SL}(3,\mathbb{R})$
2. $\mathrm{Stab}^+(\phi_2) = \mathrm{SL}(3,\mathbb{C})$
3. $\mathrm{Stab}^+(\phi_3) = \mathrm{GL}(3,\mathbb{R})\ltimes{\frak{sl}(}3,\mathbb{R})$
4. $\mathrm{Stab}^+(\phi_4) = \bigl(\mathrm{Sp}(2,\mathbb{R})\times \mathbb{R}^\ast\times \mathbb{R}^\ast\bigr)\ltimes\bigl(\mathbb{R}^4\oplus \mathbb{R}^5\bigr)$
5. $\mathrm{Stab}^+(\phi_5) = \bigl(\mathrm{SL}(3,\mathbb{R})\times\mathrm{GL}(3,\mathbb{R})\bigr)\ltimes\bigl(\mathbb{R}^3\otimes \mathbb{R}^3\bigr)$
6. $\mathrm{Stab}^+(\phi_6) = \mathrm{GL}(6,\mathbb{R})$

For a proof of this classical fact, you can see the Appendix in my article On the geometry of almost complex $6$-manifolds (available at http://arxiv.org/abs/math/0508428).