Explicitly, as a free module over the ground ring $\mathbb{M}_2$:
$$H^{\ast,\ast}(\text{Gr}_2(\mathbb{R}^{10,5}))=\mathbb{M}_2\oplus\Sigma^{1,1}\mathbb{M}_2\oplus\Sigma^{2,1}\mathbb{M}_2\oplus\Sigma^{2,2}\mathbb{M}_2\oplus\Sigma^{3,2}\mathbb{M}_2\oplus\Sigma^{4,2}\mathbb{M}_2\oplus\Sigma^{4,3}\mathbb{M}_2\oplus\Sigma^{5,3}\mathbb{M}_2\oplus\Sigma^{6,3}\mathbb{M}_2\oplus\Sigma^{6,4}\mathbb{M}_2\oplus\Sigma^{7,4}\mathbb{M}_2\oplus\Sigma^{8,4}\mathbb{M}_2\oplus\Sigma^{8,5}\mathbb{M}_2\oplus\Sigma^{9,5}\mathbb{M}_2\oplus\Sigma^{10,5}\mathbb{M}_2\oplus\Sigma^{10,6}\mathbb{M}_2\oplus\Sigma^{11,6}\mathbb{M}_2\oplus\Sigma^{12,6}\mathbb{M}_2\oplus\Sigma^{12,7}\mathbb{M}_2\oplus\Sigma^{13,7}\mathbb{M}_2\oplus\Sigma^{14,7}\mathbb{M}_2\oplus\Sigma^{14,8}\mathbb{M}_2\oplus\Sigma^{15,8}\mathbb{M}_2\oplus\Sigma^{16,8}\mathbb{M}_2.$$