If the answer is negative, I wonder under what conditions it would be semi-local.
EDIT Here's an example of a local domain which is not necessarily a Japanese ring. Let $A$ be a valuation ring, $K$ its field of fractions. Let $P$ be the maximal ideal of $A$. Let $L$ be a finite extension of $K$. Let $B$ be the integral closure of $A$ in $L$. It is well-known that there exist only finitely many valuation rings of $L$ dominating $A$. Let $M$ be a maximal ideal of $B$. It is well-known that $M$ lies over $P$. Hence $B_M$ dominates $A$. There exists a valuation ring $R$ of $L$ dominating $B_M$. Since $M$ is determined by $R$ and $R$ dominates $A$, there exist only finitely many maximal ideals of $B$. Hence $B$ is a semilocal ring.
Note that $B$ is not necessarily a finite $A$-module(even if $A$ is a discrete valuation ring). Hence $A$ is not necessarily a Japanese ring.