For Question 1, note that if $A$ is a colimit of a diagram, then $A$ is generated as a ring by the subrings of $A$ which are images of the objects in the diagram. In particular, if $A$ is a colimit of a functor on $\mathcal{J}$ with values in Noetherian rings, then $A$ is generated by at most $\kappa$ Noetherian subrings of $A$, where $\kappa$ is the cardinality of the set of objects of $\mathcal{J}$. So to show the answer is no, it suffices to give examples of rings which cannot be generated by any given number of Noetherian subrings.
To prove this, let $R=\mathbb{Z}[S]$ where $S$ is an infinite set of indeterminates. I claim that every Noetherian subring of $R$ is contained in $\mathbb{Z}[S_0]$ for some finite $S_0\subset S$. It follows that $R$ cannot be generated by fewer than $|S|$ Noetherian subrings.
To prove the claim, suppose $R_0\subseteq R$ is not contained in $\mathbb{Z}[S_0]$ for any finite $S_0\subset S$. We recursively construct a sequence of elements $(r_n)$ in $R_0$ with no constant term such that the chain of ideals $$0\subset (r_0)\subset (r_0,r_1)\subset\dots$$ is strictly increasing, to conclude that $R_0$ is not Noetherian. Having chosen $r_0,\dots,r_{n-1}$, let $s\in S$ be a variable which appears in some element of $R_0$ but does not appear in $r_0,\dots,r_{n-1}$ (such an $s$ exists by our hypothesis on $R_0$). Let $r_n$ be an element of $R_0$ which contains a monomial involving $s$ of minimal total degree (say, of degree $d$). By add an integer to $r_n$, we may assume $r_n$ has no constant term. I claim that $r_n\not\in(r_0,\dots,r_{n-1})$ and so $(r_0,\dots,r_{n-1})\subset (r_0,\dots,r_n)$ strictly. Indeed, consider an arbitrary element $$x=\sum_{i=0}^{n-1}a_ir_i\in(r_0,\dots,r_{n-1})$$ for $a_i\in R_0$. Note that every monomial in $x$ which contains $s$ has degree greater than $d$, since such a monomial in an $a_i$ must have degree at least $d$ and each $r_i$ has no constant term. Thus $x\not=r_n$, as desired.
For Question 2, the answer is yes. For instance, you could take $\mathcal{J}$ to be the disjoint union of all small categories. Since every ring is a small colimit of Noetherian rings, you can write any ring as a $\mathcal{J}$-indexed colimit of Noetherian rings (just send all the other components of $\mathcal{J}$ to $\mathbb{Z}$).
Best Answer
There's an example in
Gilmer, Robert; Heinzer, William, Artinian subrings of a commutative ring, Trans. Am. Math. Soc. 336, No. 1, 295-310 (1993). ZBL0778.13012.
of two artinian subrings of a commutative artinian ring whose intersection is not artinian. Since the intersection is the pullback of the inclusion maps, this answers the question.
Let $R=\mathbb{C}(x)[y]/(y^2)$. Then $R$ is a two-dimensional algebra over $\mathbb{C}(x)$, and is therefore artinian.
Let $R_1=\mathbb{C}(x^2)+y\mathbb{C}(x)\subset R$. Then $R_1$ is a three-dimensional algebra over $\mathbb{C}(x^2)$, and is therefore artinian.
Let $R_2=\mathbb{C}(x^2+x)+y\mathbb{C}(x)\subset R$. Then $R_2$ is a three-dimensional algebra over $\mathbb{C}(x^2+x)$, and is therefore artinian.
But $\mathbb{C}(x^2)\cap\mathbb{C}(x^2+x)=\mathbb{C}$, so $R_1\cap R_2=\mathbb{C}+y\mathbb{C}(x)$, which is not artinian, or even noetherian, since $y\mathbb{C}(x)$ is not finitely generated as an ideal.