What does a voltmeter actually measure? For time varying fields (even quasistatic ones) the electric field is given by 
$${\bf E} = - \nabla \Phi - \frac{\partial {\bf A}}{\partial t}$$
So what does a voltmeter measure? Does it measure a difference in $\Phi$ between two points $a$ and $b$, or does it measure $\int_a^b {\bf E} \cdot d{\bf l}$ or does it measure something else? 
 A: A typical voltmeter contains an internal Ohmic resistor with known and very high resistance $R$ (called the "input resistance" or "input impedance"), and an extremely sensitive ammeter that measures the current through that resistor.  When the voltmeter is connected in parallel across some circuit elements, then ideally the internal resistor has resistance so much higher than any of the circuit elements across which the voltage is being measured, that it acts as an almost perfect insulator, and the vast majority of the current (if there is any) flows across the circuit elements to be measured.  The voltmeter measures the tiny amount of current $I$ that does pass through its internal resistor, and then displays the voltage $V = IR$.  Aside from calibration error in $R$ and direct measurement error of the current, the voltmeter's displayed value will become incorrect when the voltage to be measured is so high that the internal resistor's response ceases to be Ohmic.
There's a nontrivial optimization problem in how high to make the voltmeter's input resistance: the higher it is, the less current passes through the voltmeter, which makes it more difficult to measure precisely.  On the other hand, you usually don't want an appreciable fraction of the total current to pass through the voltmeter, or else merely connecting the voltmeter to the circuit will actually change the circuit's behavior, which is usually not desired.
As you suggested in your question, it is impossible to measure $\Phi$ in general - even in principle - because it is a gauge-dependent quantity.  A real voltmeter typically consists of two wires (that are good conductors) which connect to the circuit element, with the actual measuring apparatus in between them.  For simplicity, let's idealize it to a single wire with uniform and known resistivity $\rho$, which can measure the current $J$ passing through itself at every point.  Then Ohm's law gives that the electric field $\vec{E} = \rho \vec{J}$, and so the displayed "voltage" actually physically corresponds to $\int_a^b \vec{E} \cdot d\vec{l}$ along the wire, as you suggested.  By Faraday's law, any time-dependent magnetic field will induce an electric current, and so the reading that the voltmeter displays will change as the wires connecting it to the circuit pass through the changing magnetic field.  So if there are magnetic fields nearby that are changing appreciably over time, then the readout on the voltmeter will depend on the exact position of the wires connecting it to the circuit, and in general this quantity is not very useful.
(In a more realistic model of a voltmeter where we take into account the fact that the resistance is much, much higher for the internal resistor than for the external wires, we find that $\vec{E}$ is typically negligible inside the wires and roughly constant across the internal resistor, and so the voltmeter's reading basically just equals $\vec{E} \cdot \vec{l}$, where $\vec{l}$ is the oriented length of the internal resistor and $\vec{E}$ is the electric field at its location.  In fact, most local electric field sensors have the same basic design as voltmeters: an extremely sensitive ammeter measuring across an Ohmic resistor of known length and input resistance.  But in this case you want the input resistance to be really high (typically about $10^4$ times that of a voltmeter) in order to prevent any current from flowing across it, as the resulting charge buildup would partially cancel the external field.)
A: A voltmeter does not measure $\Phi$ (the electrostatic potential), nor the difference  $\Phi_A - \Phi_B$ where $A$ and $B$ are arbitrary points in a material. It also does not measure $-\int \vec E \cdot d\vec l$.
A voltmeter measures the difference in the electrochemical potential between point $A$ and point $B$, divided by the elementary charge. This corresponds to $\Delta V = -\int (\vec E +\nabla \mu/e) \cdot d\vec l$ where $\mu$ is the chemical potential. 
In some cases, it may correspond to  $-\int \vec E \cdot d\vec l$, or it may measure exclusively $\mu/e$. The general case is a mix of both.
An example given by Apertet et al. is the one of a p-n junction at equilibrium. In that case even though there is a built-in electrostatic potential $V_\text{in}$, a voltmeter would read a nil voltage because $eV_\text{in}=\nabla \mu$ and so $\tilde{\mu}=0$ where $\tilde{\mu}$ is the electrochemical potential.
References: Solid State Physics by Ashcroft and Mermin (1976) page 257. Apertet et al., Riess and this university website.
A: Although both answers above have valid correct facts, they are both missing the original point in the question. The op tried to ask what happens in the electrodynamic case when the line integral of the electric field is path dependant.
In this case the line integral would give a different value for each trajectory of the electric field.
Above was mentioned (Walter lewins classes were used as reference) that in the end the voltmeter would measure the line integral of the electric field (the nonconservative electric field in this case).
This is correctBut it is important to mention that Walter Lewins explanation was for the voltage in a line element without volume. In the general case of a volume element, the voltmeter (at least the classic one based in a galvanometer) measures an average voltage around a cross sectional area perpendicular to all current lines.
Imagine having two cross sectional surfaces in a resistance. Then you would calculate each line integral from each origin point in the initial cross sectional surface following the current lines to the landing cross sectional surface.
There is a line integral for each point in the initial cross sectional surface, you would have to average all the line integrals over the cross sectional surface, and that is what you area measuring with a classic voltmeter. You are measuring the average work per unit charge done in the volume element you are considering, limited by the two cross sectional surfaces normal to the current lines.
