In the following argument $\mathbb{F}_p$ denotes either a finite field or the field $\mathbb{F}_0 = \mathbb{Q}$.
Yes. Let $\phi : R \to S$ be a morphism of finitely-generated $\mathbb{Z}$-algebras and let $m$ be a maximal ideal of $S$. Then $S/m$ is a finitely generated (as a ring) field.
Lemma: Finitely generated fields are finite fields.
Proof. Any finitely generated field $F$ is finitely generated over $\mathbb{F}_p$ for some $p$. By Noether normalization $F$ is a finite integral extension of $\mathbb{F}_p[x_1, ... x_n]$ for some $n$, and since it is a field we must have $n = 0$. Hence $F$ is either a finite field or a number field, but the latter is impossible as rings of integers in number fields have infinitely many primes.
It follows that the image of $R$ in $S/m$ is a subring of a finite field, hence also a finite field. Hence $m$ is sent to a maximal ideal of $R$.
Let $k$ be a field and $\bar{k}$ its algebraic closure. An algebraic zero of a subset $\Phi$ of $k[x_1,\dots,x_n]$ is an element $(a_1,\dots,a_n) \in \bar{k}^n$ such that $f(a_1,\dots,a_n)=0, \, \forall f \in \Phi$. Then Hilbert's Nullstellensatz says that if $g \in k[x_1,\dots,x_n]$ vanishes at every algebraic zero of $\Phi$, then $g$ is inside the radical of the ideal generated by $\Phi$ (Matsumura, Theorem 5.4).
The next key thing to observe is that given an ideal $I$ of $k[x_1,\dots,x_n]$, there is a $1-1$ correspondence between algebraic zeros of $I$ and maximal ideals of $k[x_1,\dots,x_n]$ that contain $I$. To see that, note that if $m$ is a maximal ideal that contains $I$ and we define $a_i$ to be the class of $x_i$ mod $m$, then $(a_1,\dots,a_n)$ is an algebraic zero of $I$ by the Zariski Lemma. Conversely, if $(a_1,\dots,a_n)$ is an algebraic zero, then $k[a_1,\dots,a_n] = k(a_1,\dots,a_n)$ and the kernel of the $k$-algebra homomorphism $k[x_1,\dots,x_n] \rightarrow k[a_1,\dots,a_n]$ that sends $x_i$ to $a_i$ is a maximal ideal (Matsumura, Theorem 5.1).
Finally, we clearly have that $I \subset \cap_{m \supset I} m$. Conversely, let $f \in \cap_{m \supset I} m$. Then $f$ vanishes at every algebraic zero of $I$ and by Hilbert's Nullstellensatz $f \in \sqrt{I}$.
Best Answer
Take $A=k[t]_{(t)}$. This is a local domain with unique maximal ideal $(t)$, so the intersection of maximal ideals containing $(0)$ is $(t)$, which is not equal to $\sqrt{(0)}=(0)$.
It is not finitely generated as a $k$-algebra: if it were, $k[t]_{(t)}[x]$ would be finitely generated as a $k$-algebra, but $k[t]_{(t)}[x]/(1-tx)\cong k(t)$, which contradicts Zariski's lemma that any quotient of a finitely generated algebra over $k$ by a maximal ideal is a finite extension of $k$.
The sort of ring that makes this second statement true is a Jacobson ring - what's going on geometrically is that for such a ring $A$, the closed points of $\operatorname{Spec} A$ are dense in every closed subset.