No, $R$ is not necessarily Japanese. What follows is a explicit example from a note that [Rankeya Datta][1] and I wrote, now [available on his website][2].

**Example.** We use Hochster's example [[Hochster][3], Ex. 1]; see [one of my other answers][4] for the relevant results therein. Let $I$ be the set of positive integers, and set
$$R_i := k[x_i^2,x_i^3] \qquad\text{and}\qquad P_i := (x_i^2,x_i^3) \subseteq R_i$$
for every $i$, where $k$ is a fixed algebraically closed field. Then, setting $R' := \bigotimes_{i \in I} R_i$, the ring
$$R := \Bigl(R' \smallsetminus \bigcup_{i \in I} P_iR'\Bigr)^{-1}R'$$
is a domain such that all of its local rings are excellent (hence Japanese), and such that the regular locus in $\operatorname{Spec}(R)$ is not open [[Hochster][3], Prop. 1].

We claim that $R$ is not Japanese. By Hochster's construction (see [[Datta–M][2], Rem. 4]), the ring $R$ is one-dimensional. But the regular locus is open for any one-dimensional Japanese domain [[Datta–M][2], Lem. 5], and hence $R$ cannot be Japanese.

  [1]: https://rankeya.people.uic.edu/
  [2]: https://rankeya.people.uic.edu/Japanese%20is%20not%20local.pdf
  [3]: https://doi.org/10.1215/S0012-7094-73-04020-9
  [4]: https://mathoverflow.net/a/339759/33088