The answer is no.

Indeed, if a martingale is a.s. everywhere differentiable, then its [quadratic variation][1] is a.s $0$. So, by the [Burkholder–Davis–Gundy inequality][2], the martingale is a.s. constant.


  [1]: https://en.wikipedia.org/wiki/Quadratic_variation#Definition
  [2]: https://en.wikipedia.org/wiki/Quadratic_variation#Martingales