Neutralisation of weak acid by base
As @Maurice pointed out, the result of a neutralization reaction is not impacted by the dissociation percentage.

Is $\pu{[}\ce{HA}{] = 10^{−3} M}$ or is $\pu{[}\ce{H+}{] = 10^{−3}}$?

$\pu{[}\ce{HA}{] = 10^{-3}}$.
Always remember, neutralization in aqueous medium depends upon the number of equivalents of acid and base:
$$N_1V_1=N_2V_2 $$  Since $V_1=V_2$, therefore, $\pu{[}\ce{HA}{]=[}\ce{NaOH}{]}$.

Why is it wrong to say that since $\ce{NaOH}$ completely dissociates, and solution is neutral, $[\ce{OH−}]$ $\pu{= 10^−3 M = [}\ce{H+}]$?

The secret lies in the name of the concept itself. Chemical equilibrium. To attain an equilibrium state, all favorable chemical reactions are first carried out. The remaining products/ions constitute the equilibrium state. Coming back to the question, the neutralisation reaction between $\ce{HA}$ and $\ce{NaOH}$ is carried out first:
$$\ce{HA + NaOH -&gt;NaA +H2O}$$
The resultant ions constitute the equilibrium, which, are none unless you consider the hydrolysis of the resultant salt of a weak acid/strong base.