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Can a galvanic cell with the overall reaction $$\ce{Cu(s) + 2H+(aq) -> Cu^2+(aq) + H2(g)}$$ work?
$\Delta G$ of this reaction is negative (because $E^{\circ}$ of the reaction is $\mathrm{0.34~V}$ as the standard electrode potential of copper will be $\mathrm{-0.34~V}$ as it is acting as an anode). This shows the cell is spontaneous;
However, I have read that $\ce{Cu(s)}$ cannot reduce $\ce{H+}$ so I was wondering if this cell would work?
TL;DR - What you know from the activity series of metals is still true. You made a mistake in your interpretation of standard electrode potential. The rationale follows:
Standard electrode potentials are written in the reduction direction:
$$\ce{M}^{x+} + x\ce{e- -> M(s)}$$
Standard electrode potentials are measured against the standard hydrogen electrode:
$$\ce{2H+ + 2e- -> H2(g)}$$
Thus for the $\ce{Cu(s)|Cu^{2+}}$ electrode, $E^\circ$ is defined for the followed redox reaction (everything at 1 molar, 1 bar, and in standard states):
$$\ce{Cu^2+ + H2 -> Cu(s) + 2H+}\ \ \ E^\circ = +0.34$$
For the oxidation of copper with acid, the electrode potential would be negative in the direction you wrote it, which translates to a positive $\Delta G^\circ$, since: $$\Delta G^\circ = - nFE^\circ$$
