The answer of Gregory Arone can be generalized to give a section for all $n\geq 3$ using a remark I learned from the writings of Fred Cohen. For $n\geq 3$ there is a homotopy equivalence
$$ SO(3)\times F_{n−3}(S^2 \text{ minus 3 points}) \simeq F_n S^2 $$
Note that $S^2$ minus three points is homeomorphic to a plane minus two points.$ %% PREAMBLE %%
\newcommand{\RR}{\mathbb{R}}
\newcommand{\identity}{\mathrm{id}}
\newcommand{\projection}{\mathrm{proj}}
\newcommand{\To}{\longrightarrow}$

Here are some details. For $n=3$, the map $SO(3) \to F_3(S^2)$ takes a linear map $L$ to the configuration $(L(1,0,0),−L(1,0,0),L(0,1,0))$ (other variations on this map are obviously possible). The case for general $n$ follows from the case $n=3$ using the action of $SO(3)$ on $S^2$. In particular, the map giving the homotopy equivalence
$$ T_n : SO(3)\times F_{n−3}(S^2 \setminus \{ (1,0,0),(-1,0,0),(0,1,0) \} ) \To F_n S^2 $$
is defined by
$$ T_n(L,(x_1,x_2,\ldots,x_{n-3})) = (L(1,0,0),-L(1,0,0),L(0,1,0),L(x_1),L(x_2),\ldots,L(x_{n-3})) $$

Further, the projections $\pi:F_n S^2 \to F_m S^2$ for $n\geq m$ correspond under the above homotopy equivalences to the map
$$ \identity_{SO(3)}\times\projection : SO(3)\times F_{n−3}(S^2 \text{ minus 3 points}) \To SO(3)\times F_{m−3}(S^2 \text{ minus 3 points}) $$
which is the product of the identity map on $SO(3)$ with the projection on configuration spaces of $S^2$ minus three points. More precisely, we have a commutative square
$$ \pi\circ T_n = T_m \circ (\identity_{SO(3)}\times\projection) $$

Consequently, the projection $\pi:F_n S^2\to F_3 S^2$ has a section. We simply need to observe that $\pi$ is a fibration, and that the following homotopy equivalent fibration
$$ SO(3)\times F_{n−3}(S^2 \text{ minus 3 points}) \To SO(3) $$
is trivial, and therefore has a section. This recovers Gregory Arone's answer, except for the neat expression he gave.

More generally, we can find a section of $\pi:F_n S^2 \to F_m S^2$ for $n\geq m$. We use the fact that $S^2$ minus three points is homeomorphic to $\RR^2$ minus two points, together with a section of the projection
$$ \projection : F_{n−3}(\RR^2 \text{ minus 2 points}) \To F_{m−3}(\RR^2 \text{ minus 2 points}) $$
constructed by adding points "near infinity" (i.e. far away from the points already in the configuration) on the positive $x$-axis. Together with the above homotopy equivalences and commutative square, we thus obtain a section of $\pi : F_n S^2 \to F_m S^2$.

[**Edit:** I corrected the following paragraph to account for Gregory Arone's first comment below. In any case, the geometric description given in my first comment below is simpler.]

Intuitively, what is the above section of $\pi$ doing? Given a configuration $(x_1,\ldots,x_m)$ of $m$ points in $S^2$, it is simply taking the first three points $x_1$, $x_2$, $x_3$, and using them to add points to the configuration — thus obtaining a configuration $(x_1,\ldots,x_m,\ldots,x_n)$ — in the manner now described. View $S^2$ as a Riemann surface and take the unique Moebius transformation (i.e. complex automorphism of the Riemann sphere) $f:S^2\to S^2$ with
$$ f(1,0,0)=x_1 \qquad f(-1,0,0)=x_2 \qquad f(0,1,0)=x_3 $$
and then place the new points $x_{m+1}, \ldots, x_n$ along the image by $f$ of the shortest geodesic from $(-1,0,0)$ to $(0,1,0)$, but very, very near to $x_2$. The neat expression given by Gregory Arone almost works, except that (it introduces an extra rotation by 90 degrees and) you have to further deform it in some way which depends on the rest of the points $x_4,\ldots,x_m$, to make sure the point you add is sufficiently close to $x_2$.

[**Later edit:** User gcousin has added a new answer to this thread which contains, in particular, an analytic expression for the section as described in the preceding paragraph.]

For completeness, I would like to add a couple of remarks for the cases $n=1$ and $n=2$, which are not interesting from the question's perspective as $F_1 S^2$ and $F_2 S^2$ are both equivalent to $S^2$ and thus are simply connected. The projection $F_2 S^2\to F_1 S^2=S^2$ is a fibration and a homotopy equivalence, and hence has a section. On the other hand, the projection $\pi:F_3 S^2\to F_2 S^2$ does not admit a section: composing it with the homotopy equivalences $T_n$ and $\pi : F_2 S^2 \to F_1 S^2 = S^2$, we get the composite map
$$ f : SO(3) \overset{T_3}{\To} F_3 S^2 \overset{\pi}{\To} F_2 S^2 \overset{\pi}{\To} S^2 $$
defined by $f(L)=L(1,0,0)$. This map $f$ is the projection onto $S^2$ of the principal $SO(2)$-bundle associated with the tangent bundle of $S^2$; since $S^2$ is not parallelizable, that bundle does not admit a section.

Unfortunately, the above answer only works for the spherical pure braid groups. In any case, perhaps the above decompositions can still give some practical geometric insight into the spherical braid groups which will help to answer the original question.