I know that $x \in$ Frac($A$), $x^n \in A$ for some integer $n>0$, does not imply that $x \in A$, as the example in the comments shows.
This leads me to the following question: is $A$ integrally closed if $x \in$ Frac($A$), $x^n \in A$ for some integer $n>0 \Rightarrow x \in A$.
For the original question: in $A = \Bbb Z[\sqrt 8]$ we have $(\sqrt 8/2)^2 = 2$ and $\,\sqrt8/2\not\in A\ \ $
For the update: let's recall some definitions that are "root" (radical) analogs of integral closure.
If $D$ is a domain with fraction-field $K$ then $D$ is $n$-root closed if $\,x\in K,\, x^n \in D\Rightarrow x\in D$
$D$ is root-closed if it is $n$-root closed for all $\,n\ge 1$.
It's not hard to determine which quadratic number rings are root closed, e.g. we have the following result, which is Corollary $2.2$ here.
Theorem $ $ A non-integrally closed algebraic order $R$ with conductor $I$ is root closed $\iff I\,$ is an intersection of maximal ideals $\,P_i\,$ of $\,\bar R = $ integral closure of $R$, such that $|\bar R/P_i| = 2$ for each $P_i$
As they remark: in the quadratic case this yields $R=\Bbb Z[\sqrt d]$ is root-closed $\iff d\,$ is square-free and either $\,d\equiv 2,3\pmod{\!4}\,$ or $\,d\equiv 1\pmod{\!8}.\,$ In the first case is $R$ integrally closed. The second case is the only non-integrally closed case satisfying the conditions of the Theorem. In fact, $\,2\,$ decomposes in $\bar R\,$ and $\,2\bar R\,$ is the conductor of $R,$ an intersection of two maximal ideals $P_i$ of $\bar R$ with $|\bar R/P_i| = 2\,$ for $i = 1, 2.$
There is much prior work on root-closure (and the closely related notion of seminormal, i.e. $\,x\in K,\, x^2,x^3\in D\Rightarrow x\in D)\,$ - much of which can be located by chasing citations starting with the linked paper.