I've just come up with the following proof(since this is an exercise, I have been trying to solve it by myself).
We use induction on the number of generators of $B$ over $A$.
It suffices to to prove the assertion when $B = A[x]$.
Let $K$ be the field of fractions of $A$.
Suppose $x$ is not algebraic over $K$.
Let $b = a_0x^n + \cdots + a_1x + a_n$, where $a_i \in A$ for $i = 0,1,\dots,n$ and $a_0 \neq 0$.
Let $\psi\colon A \rightarrow \Omega$ be a homomorphism such that $\psi(a_0) \neq 0$.
Let $\alpha$ be an element of $\Omega$ which is not a root of the polynomial
$\psi(a_0)X^n + \cdots + \psi(a_1)X + \psi(a_n)$.
There exists a unique homomorphism $\phi\colon B \rightarrow \Omega$ extending $\psi$ such that $\phi(x) = \alpha$.
Then $\phi(b) \neq 0$.
It remains to prove the asssertion when $x$ is algebraic over $K$.
Then $b$ is also algebraic over $K$.
Suppose $a_0x^n + \cdots + a_1x + a_n = 0$, where $a_i \in A$ for $i = 0,1,\dots,n$ and $a_0 \neq 0$.
Suppose $c_0b^m + \cdots + c_1b + c_m = 0$, where $c_i \in A$ for $i = 0,1,\dots,m$ and $c_0 \neq 0$
We may assume that $c_m \neq 0$.
Let $a = a_0c_m$.
Then $B_a = A_a[x]$ is integral over $A_a$, where $A_a$ is the localization of $A$ with respect to $\{1, a, a^2,\dots\}$.
Let $\psi\colon A \rightarrow \Omega$ be a homomorphism such that $\psi(a) \neq 0$.
$\psi$ is uniquely extended to a homomorphism $\psi'\colon A_a \rightarrow \Omega$.
Hence by this question, $\psi'$ can be extended to a homomorphism $\phi\colon B_a \rightarrow \Omega$.
We claim $\phi(b) \neq 0$.
Suppose $\phi(b) = 0$.
Then $\psi(c_m) = 0$.
Hence $\psi(a) = 0$.
This is a contradiction.