What reaction occurs when etching copper to create circuit boards?
Well your equation balances but that isn't the right reaction. 

First lets break this down into half cell reactions.

Oxidation reaction


  $\ce{Cu_{(s)} &lt;-&gt; Cu^{2+}_{(aq)} + 2e^{-}}$


Reduction reaction


  $\ce{H2O2 + 2 H^+ + 2e^{-} -&gt;  2 H2O_{(l)}}$


Decomposition reaction

Hydrogen peroxide spontaneously decomposes in acid solution. 


  $\ce{2H2O2 -&gt;  2H2O_{(l)} + O2_{(g)}}$


Stoichiometric reaction between copper and hydrogen peroxide

So the stoichiometric reaction between copper and hydrogen peroxide is:


  $\ce{H2O2 + 2 H^+ + Cu_{(s)} -&gt; Cu^{2+}_{(aq)} + 2 H2O_{(l)}}$


the $\ce{2 H^+}$ ions come from the vinegar. (Let's use $\ce{HOAc}$ for vinegar).


  $\ce{H2O2 + 2 HOAc + Cu_{(s)} -&gt; Cu^{2+}_{(aq)} + 2OAc^{-} + 2H2O_{(l)}}$


Sodium chloride's Role

Finally the role of the $\ce{NaCl}$ plays in the etch. The $\ce{NaCl}$ must be fairly high to drive the $\ce{Cu^{2+}}$ to a copper chloride complex. This thus reduces the "free" $\ce{Cu^{2+}}$ in solution which keeps the electrochemical potential of copper high so that the solution keeps dissolving copper. (Think of the half-cell copper reaction as if it is in a battery. The voltage of the half-cell would drop as the copper builds up.)