Are there digraphs such that any two arborescences are arc-disjoint?
Leaving aside the trivial case (graphs which only have one $r$-arborescence), this won't be possible.

Suppose $(V,E)$ is an $r$-arborescence of $(V,A)$. Then $E$ contains some (nonzero) number of arcs $(r,s_1),\ldots,(r,s_k)$ out of the root (otherwise we are in a trivial case). Now if $(V,F)$ is another arborescence, then either


$F$ also contains some $(r,s_i)$, so the arborescences are not arc-disjoint, or
$F$ contains $(r,t)$ for some $t\notin\{s_1,\ldots,s_k\}$, and for each $i$, an arc $(p_i,s_i)$ for some (not necessarily distinct) vertices $p_1,\ldots,p_k\neq r$. But then we can replace one such arc with $(r,s_i)$ - it is easy to see that $(V,F\cup\{(r,s_i)\}\setminus\{(p_i,s_i)\})$ is again an $r$-arborescence: For any vertex $v$ in the subtree of $(V,F)$ rooted at $s_i$, the only path from $r$ to $v$ is now the old one, with the prefix up to $s_i$ replaced by $(r,s_i)$; for all other vertices, the paths remain unchanged. This new arborescence is distinct from $(V,E)$ since it still contains $(r,t)$, but they are not arc-disjoint since they both contain $(r,s_i)$.