Finite extension of domains : preimage of non-zero ideal is non-zero?

Let $B$ be a commutative domain and $A \subseteq B$ a subring. Suppose that the extension $A \to B$ is finite, i.e. $B$ is finitely generated as an $A$-module. If $I \cap A = \lbrace 0 \rbrace$ for some ideal $I$ of $B$, can we conclude that $I = \lbrace 0 \rbrace$?

I know that if $I$ contains a nonzero prime ideal, the statement holds since $A \to B$ is an integral extension, whereby no two comparable prime ideals of $B$ can have the same intersection with $A$. But does the statement still hold if $I$ does not contain a nonzero prime ideal? And what if $A \to B$ is only assumed to be an integral extension?

A solution attempt based on user26857's hints:

Let $A \to B$ be an integral extension of commutative domains, let $I$ be an ideal of $B$ such that $I \cap A = \lbrace 0 \rbrace$. We claim that $I = \lbrace 0 \rbrace$. Take an element $x \in I$. As $x \in B$, $x$ is integral over $A$, whereby $x$ is the root of a minimal monic polynomial over $A$ $$ F = T^n + \sum_{i=1}^n \alpha_iT^{n-i} \in A[T] $$ with each $\alpha_i \in A$. Then we have $$ \alpha_n = -x^n - \sum_{i=1}^{n-1}\alpha_ix^{n-i} \in I \cap A $$ implying that $\alpha_n = 0$ and $T$ divides $F$. Hence there exists a $G \in A[T]$ monic such that $F = GT$ and because $B$ is an integral domain, either $x$ is a root of $T$ (and we are done) or $x$ is a root of $G$. But $F$ was chosen to be the minimal monic polynomial annihilating $x$, so $x$ cannot be a root of $G$!