I'm not sure what an "integral" noetherian ring is (a domain?), but it is impossible for any noetherian ring with connected spectrum. Indeed, if

If $P$ is a prime ideal containing $a$ then $\cap a^n$ is contained in $\cap P^n$, whose image in $A_P$ vanishes, so $\cap a^n$ has vanishing image in $A_P$ for every point $P$ of Spec($A/a$). HenceSuch $P$ exist as long as $a$ isn't the unit ideal, so if $A$ is an integral domain then $\cap a^n = 0$ in $A$ since $A \rightarrow A_P$ is injective in such cases.

In contrast, for $A = k[x,y]/(xy) = k[x] \times_k k[y]$ for a field $k$, and $$a = (x-1)A = (x-1)k[x] \times_k k[y],$$ we have $\cap a^n = yk[y] \ne 0$ and $\{a^n\}$ doesn't stabilize. Thus, the "domain" condition cannot be relaxed to mere connectedness of the spectrum (even assuming reducedness).

[${\bf{Remark}}$: An earlier version of this answer had the following bogus "proof" that $\cap a^n = 0$ assuming that Spec($A$) is merely connected, and some of the comments refer to this bogus argument. So here it the incorrect proof of that incorrect generalization, with the mistake in the reasoning identified.

"Proof": Arguing as above, since $\cap a^n$ has vanishing image in the stalk $A_P$ at each point of Spec($A/a$), it vanishes over an open neighborhood $U$ of Spec($A/a$). But this closed set. On the other hand, the ideal sheaf $J = \cap (\widetilde{a})^n$ is the unit ideal on the open set $V$ of Spec($A$) complementary to the closed set Spec($A/a$), so . Since $\cap a^n$ is the ideal of global sections of $J$, if $J$ is quasi-coherent (equivalently, coherent) then $U$ and $V$ are disjoint open sets that cover the entire space. By connectedness, it follows which would imply by connectedness that either $V$ is empty (in which case $\cap a^n = 0$ ) in $A$) or $U$ is empty (in which case $a$ is the unit ideal). "QED"

The error is that often the ideal sheaf $\cap (\widetilde{a})^n$ is not quasi-coherent (equivalently, isn't coherent), which is to say that it doesn't correspond to its ideal of global sections $\cap a^n$. This is an example of the fact that quasi-coherence can be destroyed by inverse limits, and more concretely inverse limits don't naturally commutes with tensor product or even just localization in general.

The failure of coherence for this ideal sheaf is seen rather concretely in the case $A = k[x,y]/(xy)$ with $a = (x-1)$ since the sheaf $\cap \widetilde{a}^n$ is the unit ideal sheaf away from the point $(1,0)$ whereas its ideal of global sections $\cap a^n = y k[y]$ is the ideal of the $x$-axis and so does not localize to the unit ideal at points $(c,0)$ with $c \ne 1$ (e.g., $c = 0$). This illustrates that the coherent ideal sheaf associated to $\cap a^n$ can be rather smaller than $\cap \widetilde{a}^n$.]

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I'm not sure what an "integral" noetherian ring is (a domain?), but it is impossible for any noetherian ring with connected spectrum. Indeed, if $P$ is a prime ideal containing $a$ then $\cap a^n$ is contained in $\cap P^n$, whose image in $A_P$ vanishes, so $\cap a^n$ has vanishing image in $A_P$ for every point $P$ of Spec($A/a$). Hence, $\cap a^n$ has vanishing image in an open neighborhood $U$ of Spec($A/a$). But this ideal is the unit ideal on the open set $V$ of Spec($A$) complementary to the closed set Spec($A/a$), so $U$ and $V$ are disjoint open sets that cover the entire space. By connectedness, it follows that either $V$ is empty (in which case $\cap a^n = 0$) or $U$ is empty (in which case $a$ is the unit ideal).