This is the corrected proof of the satement in the case when $F'$ is supposed to be continuous. We also assume that $F'$ never vanishes, othevise the statement is obviuous. The first four comments below are about previous versions of my proof, which were not correct. So first 4 comments are not relevant anymore to this answer (really sorry for that).
I will use slightly modified reasoning of Ilya (please read his answer. His idea is that we conisder the curve $\frac{F'}{|F'|}$ in the unit $S^{n-1}$). Then we need to prove a little lemma.
Lemma. Consider a continous curve $C$ on a sphere $S^{n-1}$ that doesn't lie inside any half-sphere parametrised by an inteval $[0,1]$. Prove that there exists an equator that intersects the curve for at least $n$ different values of the parameter.
Proof. The center of the sphere belongs to the convex hull of $C$ by the assumption of the lemma. This means that there are $n+1$ or less points on $C$ such that the simplex with vertices in these points contains the centre $0$ of the sphere. If the number of points is less than $n+1$, we are done. Suppose that the number of points is in fact $n+1$ and they are linearly inependent. We call the points $x(t_1),...,x(t_{n+1})$ where $t_i$ are odered for. Consider now the hyperplane $H$ generated by points $x(t_1),...,x(t_{n-1})$ and the center of the sphere. $H$ cuts $S^{n-1}$ in two halves. Since by constructions the convex hull of $x(t_i)$ contains $0$, the points $x(t_n), x(t_{n+1})$ are in different half spaces with respect to the plane. So this plane intersects the part of curve C contained between $t_n$, $t_{n+1}$ in some point $x(t_n')$. The lemma is proved.