Here is an overkill answer.

Let $A=k[x,xy^m]\subset R$. Then, as I said in a previous answer for a previous question of yours, $R$ is integral over $A$ and thus it is a finite module over $A$. It is torsion free of rank $m$ as a module $A$. Thus one has a resolution, $0\to P\to F\to R\to 0$ as $A$-modules, where $F$ is a free module of say rank $n$ over $A$. By Auslander-Buchsbaum, $P$ is projective (over $A$) of rank $n-m$ (in fact free by Seshadri's theorem). Tensoring by $A/xA$, we get an exact sequence $0\to P/xP\to F/xF\to B\to 0$. Since $P/xP$ is free (over $A/xA=k[xy^m]$), we see that $[B]=m\in G_0(A/xA)$. On the other hand, we have the natural map $G_0(B_{red})\to G_0(B)$, which as you noted is an isomorphism. Thus, the map $G_0(B)\to G_0(A/xA)$ is an isomorphism, since the composite $G_0(B_{red})\to G_0(A/xA)$ is. This shows the class of $B$ in $G_0(B_{red})$ is $m$.

Here is an overkill answer.

Let $A=k[x,xy^m]\subset R$. Then, as I said in a previous answer for a previous question of yours, $R$ is integral over $A$ and thus it is a finite module over $A$. It is torsion free of rank $m$ as a module over $A$. Thus one has a resolution, $0\to P\to F\to R\to 0$ as $A$-modules, where $F$ is a free module of say rank $n$ over $A$. By Auslander-Buchsbaum, $P$ is projective (over $A$) of rank $n-m$ (in fact free by Seshadri's theorem). Tensoring by $A/xA$, we get an exact sequence $0\to P/xP\to F/xF\to B\to 0$. Since $P/xP$ is free (over $A/xA=k[xy^m]$), we see that $[B]=m\in G_0(A/xA)$. On the other hand, we have the natural map $G_0(B_{red})\to G_0(B)$, which as you noted is an isomorphism. Thus, the map $G_0(B)\to G_0(A/xA)$ is an isomorphism, since the composite $G_0(B_{red})\to G_0(A/xA)$ is. This shows the class of $B$ in $G_0(B_{red})$ is $m$.