Are local rings of monoid algebras geometrically unibranch? 
Let $\mathrm{M}$ be a finitely generated submonoid of $\mathbb{Z}^{\oplus d}$ for some $d$, let $A := k[\mathrm{M}]$ be the associated monoid algebra over a field $k$, let $\mathfrak{m} \subset A$ be a maximal ideal corresponding to a $k$-point of $A$. Is the local ring $A_{\mathfrak{m}}$ geometrically unibranch, i.e. does the strict henselization $(A_{\mathfrak{m}})^{\mathrm{sh}}$ have a unique minimal prime?

Remarks/thoughts: If the monoid $\mathrm{M}$ is "saturated" (if $x \in \mathrm{M}^{\mathrm{gp}}$ is an element such that there exists $n \in \mathbb{Z}_{\ge 1}$ for which $nx \in \mathrm{M}$, then $x \in \mathrm{M}$), then $A$ is a normal domain (see [1, Proposition 3.4.1]), so $A_{\mathfrak{m}}$ and $(A_{\mathfrak{m}})^{\mathrm{sh}}$ are also normal domains (by e.g. [2, Tag 00GY] and [2, Tag 06DI] respectively).
References:
[1] Ogus, "Lectures in Logarithmic Algebraic Geometry", version of Oct 24, 2017, link
[2] Stacks Project
Keywords: singularities of toric varieties, geometrically unibranch, completion, henselization, minimal primes, semigroup
 A: Edit.  As Friedrich Knop points out, these schemes are typically not even $S2$.  I added an example at the end. (I believe it is the example that Friedrich Knop was suggesting.)
Original answer. This is not true.  Let $d$ equal $2.$  Let $n\geq 2$ be an integer that is prime to the characteristic of $k.$  Inside $\mathbb{Z}^{\oplus 2},$ consider the subsemigroup $M$ generated by the following three elements, $$a=(1,0),\ \ b=(0,1), \ \ c = (n,-n).$$  Denote by $x$ and $y$ the elements of $k[M]$ corresponding to $a$ and $b.$  Then $k[M]$ is the $k$-subalgebra of $k[x,x^{-1},y,y^{-1}]$ generated by $x,$ by $y,$ and by $z=x^n/y^n.$  This has presentation, $$k[M] = k[x,y,z]/\langle y^nz -x^n \rangle.$$  Now consider the maximal ideal $\mathfrak{m} = \langle x,y,z-1\rangle.$  Because $n$ is prime to the characteristic, the ring extension $$k[M] \to k[M][u]/\langle (u+1)^n-z \rangle,$$ is étale near $\mathfrak{m}$.  After adjoining $u$, it is clear the local ring near the maximal ideal $\mathfrak{n} = \langle x,y,z-1,u \rangle$ is not unibranch, i.e., $$k[x,y,u]/\langle (y(1+u))^n - x^n \rangle,$$ is not unibranch near $\langle x,y,u\rangle.$
I suspect that for every subsemigroup $M$ of $\mathbb{Z}^{\oplus d},$ the semigroup ring $k[M]$ is $S2.$  If the ring $k[M]$ is $S2,$ then examples such as the one above would be the only examples.  Precisely, if $k[M]$ is $S2$ and if $\text{Spec}\ k[M]$ is unibranch at all codimension $1$ points, then by Hartshorne's Connectedness Theorem, the scheme is unibranch everywhere.
Edit. As Friedrich Knop points out, these schemes are typically not even $S2.$  For instance, let $M$ be the subsemigroup of $\mathbb{Z}_{\geq 0}^{\oplus 3}$ with the following generators, $$M = \mathbb{Z}_{\geq 0}\cdot (1,0,0) + \mathbb{Z}_{\geq 0}\cdot (0,1,0) + \mathbb{Z}_{\geq 0} \cdot (1,0,1) + \mathbb{Z}_{\geq 0}\cdot (0,1,1) + \mathbb{Z}_{\geq 0} \cdot (0,0,2).$$  Denote the respective generators of $k[M]$ as follows, $$x = \chi^{(1,0,0)}, \ y = \chi^{(0,1,0)}, \ u = \chi^{(1,0,1)}, \ v= \chi^{(0,1,1)}, \ w = \chi^{(0,0,2)}.$$ Then a presentation for the semigroup ring is, $$k[M] = k[x,y,u,v,w]/\langle u^2 - x^2w, v^2-y^2w,xv-yu,uv-xyw \rangle.$$  This is smooth if we invert either $x$, so that the fraction $z=ux^{-1}$ is in the fraction ring, or if we invert $y$, so that the fraction $z=vy^{-1}$ is in the fraction ring.  Thus, the singular locus is contained in the common vanishing set of $x$ and $y$.  Set-theoretically, this implies that also $u$ and $v$ equal $0$, so that the singular locus is in the affine line $\text{Spec}\ k[w].$  
Since $\text{Spec}\ k[M]$ has dimension $3$ and the singular set has dimension $1$, this ring is $R1$.  Yet it is not normal, since $z$ satisfies the monic polynomial $z^2 -w=0$.  Therefore, by Serre's Criterion for normality, the semigroup ring $k[M]$ is not $S2.$
Please note, if you do the similar analysis as in the original example for the maximal ideal $\mathfrak{m} = \langle x,y,u,v,w-1\rangle,$ it turns out that this new example is even worse.  The strict henselization is not unibranch, and it is even disconnected by removing the codimension $2$ zero scheme of $\langle x,y,u,v\rangle.$  So it appears that we should expect no form of unibranchedness / connectedness for the strict henselizations of a non-normal semigroup ring.
