*Sorry*: I was being a dim-wit.

(My answer now reflects a fair amount of editing -- sorry.)

There are only finitely many isomorphism classes of $\mathbf{Q}$-forms of simple groups of type $G_2$. I expect that means
there are only finitely many iso classes of $\mathbf{Q}$-forms of simple Lie algebras of type $G_2$ though I confess I didn't carefully think through the transition from algebraic groups to Lie algebras.

Indeed, over any field $k$, $k$-forms of a simple algebraic group of
type $G_2$ are classified by the cohomology set
$H^1(k,H)$ where $H$ is a split group of type $G_2$. [We use here that $H$ is simple -- i.e. adjoint -- and that every $k$-automorphism of $H$ is inner].

Now, one knows that $H^1(\mathbf{Q}_p,H)$ is trivial for all primes p
(since $\mathbf{Q}_p$ is a local field and $G$ is simply connected). And $H^1(\mathbf{R},H)$ is finite [there are only finitely many (two, I think?) real forms]. Hence
the Hasse principle implies that $H^1(\mathbf{Q},H)$ is finite.

For a general field $k$, note that the cohomology
set $H^1(k,G)$ identifies with
(1) the set of isomorphism classes of octonion algebras over $k$, and
(2) the set of isomorphism classes of certain quadratic forms known as 3-Pfister forms. For all this, see e.g. Serre's part of the book [Garibaldi, Merkurjev, Serre "Cohomological Invariants and Galois cohomology".

At least if the characteristic of $k$ is 0 and $H_1$ and $H_2$ are two $k$-forms of the group $G_2$, I hope that $H_1 \not \simeq H_2$ should imply that $\operatorname{Lie}(H_1) \not \simeq \operatorname{Lie}(H_2)$.

So to decide if there are infinitely many $k$-forms
of the simple Lie algebra of type $G_2$, it is enough to exhibit infinitely
many 3-Pfister forms over $k$ which are not isometric.

Now, Theorem 18.1 in [Serre, loc. cit.] calculates the group of cohomological
invariants Inv$_k($Pfister$_3,\mathbf{Z}/2\mathbf{Z})$; it is a free
module of rank 2 over the (infinite) cohomology ring $H^\bullet(k,\mathbf{Z}/2\mathbf{Z})$. And there is an invariant e such that for $\alpha_1,\alpha_2,\alpha_3 \in k^\times$, the value of e on the $3$-Pfister form $Q_\alpha = \langle \langle \alpha_1$, $\alpha_2$, $\alpha_3 \rangle \rangle$ is the cup-product $(\alpha_1)\cup (\alpha_2) \cup (\alpha_3)$ in $H^3(k,\mathbf{Z}/2\mathbf{Z})$ of the classes in $H^1(k,\mathbf{Z}/2\mathbf{Z}) = k^\times/k^{\times 2}$ determined by the $\alpha_i$.
In fact, any "normalized invariant" is a $H^\bullet(k,\mathbf{Z}/2\mathbf{Z})$-multiple of $e$.

So to use these invariants to find lots of non-isometric 3-Pfister forms, you'd need at least that $H^3(k,\mathbf{Z}/2\mathbf{Z})$ is non-zero.

Now, $H^3$ is non-zero for $k=\mathbf{Q}((T))$ (or even $\mathbf{Q}_p((T))$)
and I believe I would expect there to be infinitely many isometry classes
of Pfister forms in those cases (but I'd be interested in seeing an argument...)