I think the answer is no if you stick to noetherian rings: Assume $f\colon \mathbb{Q} \to S$ is formally etale. Let $K_1, \dotsc, K_n$ denote the fraction fields of $S$ (= localizations at generic points of the irreducible components; note that $S$ is reduced since $f$ is regular (Stacks 07EL)).
Now since localizations are ind-etale the maps $\mathbb{Q} \to K_i$ are formally etale. Using Stacks 00UO $$\Omega^1_{K_i/\mathbb{Q}} = 0$$ and so $K_i/\mathbb{Q}$ is separable algebraic. In particular $\mathbb{Q} \to K_1 \times \dotsb \times K_n$ is integral. This implies that $S \to K_1 \times \dotsb \times K_n$ is also integral. Hence, $S$ is also zero-dimensional so coincides with $K_1 \times \dotsb \times K_n$.
I don't know about the non-noetherian case but generally some more pathologies can occur then (see for instance, Mukhopadhyay and Smith - Reducedness of formally unramified algebras over fields for an example of a non-reduced formally unramified $\mathbb{Q}$-algebra). Also I realize they also prove what I said above (with weaker assumptions) in Corollary 3.3 of that paper (formal unramifiedness is enough).
2022/9/21 Edit: There is a paper on the arxiv now which answers the question negatively if one also assumes $A$ reduced: Mondal and Mukhopadhyay - Ind-étale vs Formally étale
Theorem 1.3 of the paper says
Let $k$ be a field of characteristic zero and $A$ be a reduced $k$-algebra – not assumed to be noetherian. If $\Omega_{A/k} = 0$, then $A$ is ind-etale.
Note that if $k \to A$ is formally etale, then $\Omega_{A/k} = 0$ (Stacks 04FE).
