Dear Li, first of all I think that when you write "... such that $q \cap B=p$, and $q$ is the minimal such ideal in the sense of inclusion", you mean "... and $q$ is a minimal ideal...".
The answer to your question is given, I think, by the following more general result
Theorem Let $\phi: B\to A$ be a morphism of noetherian rings and ${\frak q} \subset A$ a prime ideal with inverse image ${\frak p}\subset B$. Then we have the following formula
$$dim A_{\frak q}\leq dim B_{\frak p} +dim (A_{\frak q}\otimes_B \kappa(\frak p)) $$
Notice that there is no mention of injectivity for $\phi$ , nor of a field nor of finite generation of $A$ or $B$.
Now, if $\frak q$ happens to be to be -as in your case- the generic point of one of the irreducible components of the fibre at $\frak p$ of the morphism $Spec(\phi): Spec(A) \to Spec(B)$, then the local ring of the fiber at $\frak q$, namely $A_{\frak q}/{\frak p}A_{\frak q} = A_{\frak q}\otimes_B \kappa(\frak p)$, is zero-dimensional (see reminder below) and you get the formula you wished.
Reminder The local ring of the generic point of an irreducible scheme is a ring having only one prime ideal (its nilpotent radical) and thus has dimension zero. If the scheme is also reduced, the local ring of its generic point is a field.
Bibliography
Matsumura, Commutative Algebra, Theorem 19, page 79
Matsumura, Commutative Ring Theory, Theorem 15.1, page 116
An example In view of Li's comment it might be of interest to some users to see an example.
Let $k$ be a field , $B=k[t]$ and $A=k[t,X,Y]/(t-XY)=k[t,x,y]$. Let $\phi:B\to A$ be the inclusion. Then the fibre of ${\frak p} =(t)\in Spec(B)$ is the subscheme
$F=V(t) \subset Spec(A)$. Please note that, even though $A$ and $B$ are domains, the fibre has two irreducible components with generic points ${\frak q}=(t,x)$ and ${\frak q}'=(t,y)$. The potentially confusing fact is that to the "physical" point $Q={\frak q}$ (say) are associated two local ring. On the one hand the local ring of $Q$ in the scheme $Spec(A)$, which is $ \mathcal O_{Spec(A), Q}=A_{{\frak q}}$ $=A_{(t,x)}$ . And on the other the local ring of $Q$ in the scheme $F$, which is $\mathcal O_{F,Q}=A_{{\frak q}}/tA_{{\frak q}}=k(y)$, a field as expected.