I think that this works.
EDIT: No, it doesn't. See John Rognes's comment.
Notation: For a point $x\in EG$ we may symbolically write $x=\sum_{j=0}^nt_jg_j$, where $g_0,\dots ,g_n$ is an ordered tuple of distinct elements of $G$ and $t_j\ge 0$ and $\sum_jt_j=1$. Here the face relations are accounted for by agreeing that if $t_j=0$ for some $j$ then $g_j$ can be omitted from the tuple and the term $t_jg_j$ can be omitted from the expression, without changing the point $x$.
Invariant open cover of $EG$: For $p\ge 0$ let $U_p\subset EG$ consist of the points $x=\sum_{j=0}^nt_jg_j$ such that there is a (necessarily unique) subset $S(x)\subset \{0,n\}$ of cardinality $p+1$ such that
(1) $t_j>t_k$ whenever $j\in S(x)$ and $k\notin S(x)$, and
(2) $t_j>0$ whenever $j\in S(x)$.
Thus for each $n$-simplex $\sigma=[g_0,\dots ,g_n]$ the set $U_p\cap\sigma$ is as follows. If $p\le n$ then it is the disjoint union of convex open subsets, one for each $p$-dimensional face $\tau\subset\sigma$, namely the open star, in the barycentric subdivision of $\sigma$, of the barycenter of $\tau$. It is empty if $p>n$.
This is well-defined (compatible with the face relations), and open. The union of the sets $U_p$ is all of $EG$. For each $p$ the set $U_p$ is invariant under the action of $G$.
Local trivializations: Now let's show that the bundle is trivial over the image of $U_p$, by describing a cross-section. Let $V_p\subset U_p$ consist of those points $x=\sum_jt_jg_j\in U_p$ such that when $j_0$ is the smallest element of the set $S(x)$ then $g_j=e$. This is open in $U_p$, and every point in $U_p$ is uniquely expressible as $gx$ for some $g\in G$ and $x\in V_p$.I think