I do not think so.

You can start from a Hirzebruch surface $\mathbb{F}_n$ and blow-up the same fibre three times as in the picture.

Then, by [Artin's contractibility criterion][1], you can contract the $(-2)$-curve and the $(-3)$ curve to a singular projective surface $X$, endowed with a morphism $X \to \mathbb{P}^1$ having a unique singular fibre, which is the union of two smooth $\mathbb{P}^1$ intersecting at one point. 

The singularities of $X$ (which lie on this fibre) are an ordinary node and a $1/3(1, \, 1)$-singularity. The latter, denoted by $P_2$ in the picture and coming from the contraction of the $(-3)$-curve, is not a rational double point.

[![enter image description here][2]][2]


  [1]: https://mathoverflow.net/questions/123375/contracting-a-curve-of-negative-self-intersection-on-a-surface
  [2]: https://i.sstatic.net/EY3j2.jpg