Great question. An ideal voltmeter with infinite series resistance would probably measure a non-zero potential difference. However, a real voltmeter will have a large (but finite) series resistance, so what you would actually see is a potential difference that is probably non-zero initially, but that decays to zero roughly exponentially.
First, picture the circuits without the voltmeter. The two circuits are coupled via parasitic capacitances. In general, depending on a number of factors including the net charge on each circuit, the battery voltages, the resistor values and the geometry of the circuits, there would probably be a non-zero potential difference $V_0$ between the two wires you would hook up the voltmeter to. It is not trivial to estimate what this potential difference might be: you would need to solve this electrostatics problem considering all the factors mentioned above.
Now consider what happens when you connect the voltmeter. You can model the effect of the parasitic capacitances between the two wires as a lumped capacitance $C$ in parallel with the voltmeter, where the voltmeter has a series resistance of $R_V$. As soon as as you connect the voltmeter at time $t=0$, the capacitance $C$ will begin to discharge through the voltmeter. The time constant associated with this discharge is $\tau=R_VC$, so the voltmeter will measure a voltage given roughly by
Note that after you wait long enough, the current through the voltmeter must be zero, because otherwise there would have to be a steady non-zero current through the voltmeter, which violates Kirchhoff's current law (charges would pile up in both circuits). So eventually the voltmeter will measure zero.