Are the pure genus zero mapping class groups residually torsion-free nilpotent?
They are
-a quotient of the pure braid groups (which are residually torsion-free nilpotent).
-torsion-free.
The answer is yes. This is because these groups are fundamental groups of complements of fiber-type hyperplane arrangements; the fact that such groups are residually torsion free nilpotent goes back to Falk and Randell.
Indeed we are considering the fundamental group of the moduli space $M_{0,n}$ of genus zero Riemann surfaces with $n$ distinct ordered marked points. Since the group of Möbius transformations acts 3-transitively on the Riemann sphere, we can take the first three markings to be $0,1,\infty$. This identifies $M_{0,n}$ with the space $$\{(x_1,\ldots,x_{n-3}) \in \mathbf C^{n-3} : x_i \neq x_j, x_i \neq 0, x_i \neq 1 \text{ for all }i,j\},$$ which is of the required form.