The answer is no. First observe that there are curves in $\textbf{R}^2$ with exactly $4$ critical points of curvature which self-intersect an unlimited number of times, as shown for instance in the picture below. 
[![enter image description here][1]][1]
Pull this curve to $\mathbf{S^2}$ via stereographic projection, which will preserve critical points of curvature. So we obtain a spherical curve whose torsion changes sign exactly $4$ times. A perturbation yields a simple closed curve in $\mathbf{R}^3$ which has exactly $4$ vertices but intersects some plane more than $4$ times.  



  [1]: https://i.sstatic.net/XQD1gscg.png