Reaction of Alcohol and Metal Hydroxide
In this simple case it is pretty obvious that a hydride ion is a much better base than it is a nucleophile. But that doesn’t answer everything; we could consider cases with much better nucleophiles ($\ce{Br-}$).

Another reason is that hydroxide is an even lousier leaving group than hydride is a nucleophile. It is literally impossible to do $\mathrm{S_N}$ displacements of uncharged oxygen atoms unless you are opening an epoxide (in which case you lose ring strain making the overall process favourable. It does not matter whether $\mathrm{S_N1}$ or $\mathrm{S_N2}$; $\ce{RO-}$ will never leave. To remove an oxygen atom, you always need to use strong acids that can protonate it, giving the far better leaving group $\ce{ROH+-R}$ which gives the free alcohol/water when displaced. Thus, the following reaction works well due to the very strong hydroiodic acid (I tried it):

$$\ce{R-CH2-OH + HI -&gt; R-CH2-OH2+ + I- -&gt; R-CH2-I + H2O}$$