This turns out to be a remarkably interesting question. I don't have a complete answer but here is a start.

First, for any space $X$ I'll write $F_n(X)\subset X^n$ for the space distinct $n$-tuples. The question asks whether there exist $\Sigma_n$-equivariant maps $f:F_n(B^2)\to\Delta_{n-1}$ with the fixed point $b=(1/n,\dotsc,1/n)$ not in the image. If there is such a map then we can push it away from $b$ to the boundary of $\Delta_{n-1}$, which is homeomorphic to $S^{n-2}$. On the other side, there is an obvious embedding $i:B\to\mathbb{R}^2$ and one can choose an embedding $j:\mathbb{R}^2\to B$ such that $ij$ and $ji$ are isotopic to the respective identity maps; using this we see that $F_n(B^2)$ is equivariantly homotopy equivalent to the space $X=F_n(\mathbb{R}^2)$. This space is well-known: the following paper is one entry point to the literature:

```
\bib{MR1344842}{article}{
author={Cohen, F. R.},
title={On configuration spaces, their homology, and Lie algebras},
journal={J. Pure Appl. Algebra},
volume={100},
date={1995},
number={1-3},
pages={19--42},
issn={0022-4049},
review={\MR{1344842 (96d:55005)}},
doi={10.1016/0022-4049(95)00054-Z},
}
```

In particular:

$\pi_1(X)$ is the pure braid group $Br_n$ on $n$ strings. Moreover, the higher homotopy groups are trivial, so $X$ is the classifying space $BBr_n$.

$X$ has an equivariant deformation retract $X_0$ that is a finite simplicial complex of dimension $n-1$. This means that $H^k(X)=0$ for $k>n-1$.

The cohomology of $X$ is completely known, together with the action of $\Sigma_n$. In particular, the top group $H^{n-1}(X)$ is the module known as $\text{Lie}(n)$ (or maybe the dual of that?). As a $\mathbb{Z}[\Sigma_{n-1}]$-module this is free of rank one, but the $\Sigma_n$-action is harder to describe. The standard description also implies that all Steenrod operations in $H^*(X_0;\mathbb{Z}/p)$ are trivial.

If we can show that there is no $\Sigma_n$-equivariant map from $X_0$ to $S^{n-2}$ then we will be done.

In the case $n=2$ we just have $X_0=S^1$ and $S^{n-2}=S^0$ with $\Sigma_2$ acting antipodally on both sides: it is clear that there is no equivariant map, as required.

In the case $n=3$ we have $S^{n-2}=S^1=K(\mathbb{Z},1)$, so the nonequivariant mapping set is $[X_0,S^1]=H^1(X_0)$, and one can check that this is just $\mathbb{Z}^3$ with the action given by permuting the coordinates and multiplying by the signature. The only fixed point for this action is zero, so any map $X_0\to S^1$ that is equivariant-up-to-homotopy is nonequivariantly homotopic to a constant map. Here the action of $\Sigma_3$ on $S^1$ is generated by a reflection and a rotation through $2\pi/3$, so there are no fixed points. This means that constant maps $X_0\to S^1$, although equivariant-up-to-homotopy, cannot be equivariant on the nose. I suspect that there are no equivariant maps, and it should be possible to prove this by equivariant obstruction theory (ie working up the skeleta of $X_0$) but I do not see the details at the moment.

For $n>3$ we still have an evident map $[X_0,S^{n-2}]\to H^{n-2}(X_0)$, but it need not be bijective. We can compare $S^{n-2}$ with the fibre of the map $Sq^2:K(\mathbb{Z},n-2)\to K(\mathbb{Z}/2,n)$, recalling that $Sq^2$ acts trivially on $H^*(X_0)$, which should give an explicit description of $[X_0,S^{n-2}]$. With a bit of representation theory we should be able to calculate the group of equivariant-up-to-homotopy maps $X_0\to S^{n-2}$. We would then need some equivariant obstruction theory to improve this to understand whether there are any equivariant maps. Because $\Sigma_n$ acts freely on $X_0$ and $S^{n-2}$ is nonequivariantly $(n-3)$-connected and $X_0$ is $(n-1)$-dimensional, this obstruction theory will only involve the last two or three skeleta of $X_0$, so it should hopefully be tractable.