The answer is 'no', but seeing this depends on whether your 'standard half-dimensional subspace' $\mathbb{R}^{2n}\times \lbrace0\rbrace$ is symplectic or not as a subspace of $\mathbb{R}^{4n}$. (You didn't specify the symplectic structure on the ambient space).

For example, when $n=1$, if $\mathbb{R}^{2}\times \lbrace0\rbrace\subset \mathbb{R}^{4}$ is Lagrangian, then the subspace you mention is not contractible, it's not even connected. You can see this as follows: If $e_1$ and $e_2$ are a basis of the given subspace and $\Omega(e_1,e_2)=0$, then we can complete them to a basis $e_1,e_2,e_3,e_4$ such that
$$
\Omega(e_1,e_3)=-\Omega(e_3,e_1)=\Omega(e_2,e_4)=-\Omega(e_4,e_2)=1
$$
and all other $\Omega(e_i,e_j)=0$. If $S\subset\mathbb{R}^4$ is a subspace such that it is independent from the given $2$-dimensional subspace, then it has a unique basis of the form
$v_3=e_3+p_{31}e_1+p_{32}e_2$ and $v_4=e_3+p_{41}e_1+p_{42}e_2$. We then compute that
$$
\Omega(v_3,v_4) = p_{32}-p_{41}
$$
and the condition for $S$ to be symplectic is just that $p_{32}-p_{41}\not=0$, so the set of such $S$ is not even connected.

On the other hand, if $\mathbb{R}^{2}\times \lbrace0\rbrace\subset \mathbb{R}^{4}$ is symplectic, then we can choose a basis $e_1,e_2,e_3,e_4$ of $\mathbb{R}^4$ such that $e_1$ and $e_2$ span the given subspace and
$$
\Omega(e_1,e_2)=-\Omega(e_2,e_1)=\Omega(e_3,e_4)=-\Omega(e_4,e_3)=1
$$
and all other $\Omega(e_i,e_j)=0$. Now, if If $S\subset\mathbb{R}^4$ is a subspace such that it is independent from the given $2$-dimensional subspace, then it has a unique basis of the form
$v_3=e_3+p_{31}e_1+p_{32}e_2$ and $v_4=e_3+p_{41}e_1+p_{42}e_2$. We then compute that
$$
\Omega(v_3,v_4) = 1 + p_{31}p_{42}-p_{41}p_{32}
$$
and the condition for $S$ to be symplectic is just that $1 + p_{31}p_{42}-p_{41}p_{32}\not=0$, so the set of such $S$ is diffeomorphic to the complement in $\mathbb{R}^4$ of a copy of $\mathrm{SL}(2,\mathbb{R})$. Again, this is not a connected set.