Yes. Let $W$ be the dual vector space to $V$. For $w \in W$ and $\alpha \in \mathcal{G}(V)$, define $w(\alpha)$ by $w(v_1 \wedge v_2 \wedge \cdots \wedge v_k) = \sum_{i=1}^k (-1)^{i-1} w(v_i) \ {\Large (}v_1 \wedge v_2 \cdots \wedge v_{i-1} \wedge v_{i+1} \wedge \cdots \wedge v_k {\Large )}$ and extending linearly to $\mathcal{G}(V)$. (Exercise: This is well defined.)
Now, fix $\beta \in \mathcal{G}(V)$. Define $W' \subseteq W$ by $W' = \{ w \in W : w(\beta) =0 \}.$ Define $V' = \{ v \in V : w(v)=0 \ \forall w \in W' \}.$
I claim that $\beta$ is in $\mathcal{G}(V'')$ if and only if $V''$ contains $V'$.
The easy direction: Suppose that $\beta \in \mathcal{G}(V'')$. Let $W'' = \{ w \in W : w(v)=0 \ \forall v \in V'' \}$. Then we immediately have $w(\beta) =0$ for $w \in W''$, so $W'' \subseteq W'$. Duality reverses containments, so $V'' \supseteq V'$.
The hard direction: Let $\beta$ and $V'$ be as above. We must show that $\beta \in \mathcal{G}(V')$. Choose a basis $e_1$, $e_2$, ..., $e_n$ for $V$ such that $e_{k+1}$, $e_{k+2}$, ..., $e_n$ is a basis for $V'$. For $I = \{ i_1, i_2, \ldots, i_r \} \subseteq \{ 1,2,\ldots, n \}$, define $e_I = e_{i_1} \wedge e_{i_2} \wedge \cdots \wedge e_{i_r}$, with $i_1 < i_2 < \cdots < i_r$. Write $\beta$ in this basis as $\beta = \sum_{I \subseteq \{ 1,2,\ldots, n\}} \beta_I e_I.$
Let $f_1$, $f_2$, ..., $f_n$ be the basis of $W$ dual to $V$. So $f_1$, $f_2$, ..., $f_k$ is a basis for $W'$. We have $f_i(\beta) = \sum_{i \in I} \pm \beta_I e_{I \setminus i}$ where the rule for which sign to use is unimportant.
For $i \leq k$, we have $f_i(\beta) =0$ since $f_i \in W'$. So $\beta_I=0$ whenever $I$ contains such an $i$. In otherwords, the only nonzero terms in $\beta$ are of the form $\beta_I e_I$ when $I \subseteq \{ k+1, k+2, ..., n \}$. In other words, $\beta \in \mathcal{G}(V')$. QED
Making $W$ act on $\mathcal{G}(V)$ in this way is a standard tool. It is the skew commutative analogue of $W$ acting by derivations on the ring $\mathrm{Sym}^{\bullet}(V)$.