This is false. The easiest counterexample I could come up with is the following "affine line with embedded points at every closed point":
Example. Let $k$ be an infinite field, let $R = k[x]$, and for each $\alpha \in k$ let $R_\alpha = R[y_\alpha]/((x-\alpha)y_\alpha,y_\alpha^2)$. Then $R_\alpha$ is an affine line with an embedded prime $\mathfrak p_\alpha = (x-\alpha,y_\alpha)$ at $x = \alpha$, sticking out in the $y_\alpha$-direction. Finally, let $$R_\infty = \bigotimes_{\alpha \in k} R_\alpha = \operatorname*{colim}_{\substack{\longrightarrow\\I \subseteq k\\\text{finite}}} \bigotimes_{\alpha \in I} R_\alpha$$ be their tensor product over $R$ (not over $k$); that is $$R_\infty = \frac{k[x]\left[\{y_\alpha\}_{\alpha \in k}\right]}{\sum_{\alpha \in k}((x-\alpha)y_\alpha, y_\alpha^2)}.$$ This is not a Noetherian ring, because the radical $\mathfrak r = (\{y_\alpha\}_{\alpha \in k})$ is not finitely generated. But $\operatorname{Spec} R_\infty$ agrees as a topological space with $\operatorname{Spec} R_\infty^{\operatorname{red}} = \mathbb A^1_k$, hence $|\!\operatorname{Spec} R_\infty|$ is a Noetherian topological space.
On the other hand, the map $R \to R_\alpha$ is an isomorphism away from $\alpha$, and similarly $R_\alpha \to R_\infty$ induces isomorphisms on the stalks at $\alpha$. Thus, the stalk $(R_\infty)_{\mathfrak q_\alpha} = (R_\alpha)_{\mathfrak p_\alpha}$ at $\mathfrak q_\alpha = \mathfrak p_\alpha R_\infty + \mathfrak r$ is Noetherian. Similarly, the stalk at the generic point $\mathfrak r$ is just $R_{(0)} = k(x)$. Thus, we conclude that all the stalks of $R_\infty$ are Noetherian. $\square$