The compact simple Lie group $\text{G}_2$ has a well-known faithful representation on $\mathbb{R}^7$, so we can regard $\text{G}_2 \leq \text{GL}_7(\mathbb{R})$. Equipping $\mathbb{R}^7$ with the usual euclidean inner product, it turns out that $\text{G}_2 \leq \text{SO}(7)$.
My question is that in the title: Can we write $\text{G}_2 = \text{SO}(7) \cap H$, where $H \neq \text{G}_2$ is also a subgroup of $\text{SL}_7(\mathbb{R})$?
I would be very happy to hear that the answer is "yes," but strongly suspect that the answer is "no."
Addendum: More generally, I'd be interested in learning the existence of connected Lie subgroups $K$ with $\text{G}_2 < K < \text{SL}_7(\mathbb{R})$ other than $K = \text{SO}(7)$.
This isn't quite a full answer, but it does answer your addendum (where you have added the word 'connected').
Under the action of $G_2$, the Lie algebra $\mathfrak{sl}_7$ decomposes as $\mathfrak{g}_2 \oplus \mathbb{R}^7 \oplus S^2_0(\mathbb{R}^7)$. Now, $\mathfrak{so}(7) = \mathfrak{g}_2 \oplus \mathbb{R}^7,$ so the only candidates for $\mathfrak{h}$ are $\mathfrak{g}_2$ and $\mathfrak{g}_2 \oplus S^2_0(\mathbb{R}^7).$ But $\mathfrak{g}_2 \oplus S^2_0(\mathbb{R}^7)$ is not closed under the Lie bracket. For example, $ [ E_{1,1} - E_{2,2}, E_{1,2}+E_{2,1} ] = 2 E_{1,2}-2 E_{2,1}$, which is not an element of $\mathfrak{g}_2 \oplus S^2_0(\mathbb{R}^7)$ (I'm using the standard definition of $G_2$ here of course). This shows that the identity component of $H$ is in fact $G_2$.