The reaction will, in fact occur. The correct formula would be:
NaOH + R-OH --> R-ONa + H2O
or if you prefer the net ionic equation:
OH- + R-OH --> R-O- + H2O
where R-OH represents the alcohol. So, in this case, it's the hydroxide reacting with the alcohol, not the sodium ions. The sodium ions are just a spectator in the reaction.
The free energy difference between the products and reactants side of the equation is not very large, however, so there will be some equilibrium between hydroxide ions (OH-) and alkoxide ions (R-O-) in solution, which will depend on the pKa of the alcohol.