Here is an (hopefully correct) example of two commutative noetherian rings which do not have a coproduct in $\textrm{Noeth}$.
Assume that the coproduct $C=\overline{\mathbb{Q}} \sqcup \overline{\mathbb{Q}}$ exists in $\textrm{Noeth}$. Letting $A=\overline{\mathbb{Q}} \otimes_{\mathbb{Z}} \overline{\mathbb{Q}}$, we then have a canonical ring map $\varphi : A \to C$. Now the idea is to consider some kind of completion of $A$, similar to the one you suggest in the last paragraph of your question: for every $\sigma \in G=\mathrm{Gal}(\overline{\mathbb{Q}}/\mathbb{Q})$, there is a noetherian quotient $\mu_\sigma : A \to \overline{\mathbb{Q}}$ given by $\mu_\sigma(x \otimes y)=x \sigma(y)$. Moreover, the resulting morphism $i : A \to \prod_G \overline{\mathbb{Q}}$ is injective (this can be checked by restricting to $K \otimes K$ where $K$ is any finite Galois extension of $\mathbb{Q}$). Applying the coproduct property in $\textrm{Noeth}$, each $\mu_\sigma$ factors through $C$, so that $i$ factors through $\varphi$. In particular, the map $\varphi$ is injective, and we may identify $A$ with a subring of $C$.
Since $A$ is integral over $\overline{\mathbb{Q}}$, it has Krull dimension $0$. In particular, every maximal ideal $\mathfrak{m}_\sigma=\ker \mu_\sigma$ is a minimal prime ideal of $A$. By a result in Bourbaki, Commutative algebra (Chap. 2, Sect. 2.6, Prop. 16), the ideal $\mathfrak{m}_\sigma$ lies above a minimal prime ideal $\mathfrak{p}_\sigma$ of $C$. So we have constructed infinitely many minimal prime ideals of $C$, which is absurd because $C$ is noetherian.