What does the $I$-$V$ curve in josephson junction mean? According to the $I$-$V$ curve for Josephson junction tunneling for S-I-S (superconductor-insulator-superconductor),

Do we have any tunneling current for $0< V\leq V_c$? If yes, then why don't we show it in the diagram? If no, why don't we have a tunneling current in this voltage interval?
 A: 
Do we have any tunneling current for $0 < V \leq V_c$?

Yes.

If yes, then why don't we show it in the diagram?

It is in the diagram, you just have to understand how that diagram was measured.
Junction basics
The Josephson junction is governed by two equations
\begin{align}
I &= I_c \sin(\delta) \\
V &= (\Phi_0 / 2\pi) \dot{\delta}
\end{align}
where $\delta$ is the superconducting phase difference across the junction, $I$ is the current through the junction, $I_c$ is the junction's critical current (determined by internal structure), $V$ is the voltage difference across the junction, and $\Phi_0 \approx 2 \times 10^{-15} \, \text{Weber}$ is the flux quantum.
A junction, being typically composed of two superconducting electrodes separated by an insulating barrier, also has some capacitance $C$.
If the critical current of the junction is exceeded then quasiparticle excitations are generated.
The quasiparticle current is not superconducting and is lossy like a normal metal current, so we represent this as a resistor $R$.
This gives us the resistively and capacitivly shunted junction (RCSJ) model of the Josephson junction, as shown in Figure 1.

Figure 1: The junction is modelled as an ideal tunneling element (the cross) in parallel with a capacitor $C$ and resistor $R$.
We also include a current bias $I$, and for the sake of connecting to real experiment we show an amplifier used to measure the small voltages involved in Josephson junction experiments.
Writing out Kirchov's circuit laws for the RCSJ model you can find
$$\frac{\Phi_0}{2\pi} C \ddot{\delta} + \frac{\Phi_0}{2\pi R} \dot{\delta} = I - I_c \sin(\delta) \, .$$
Rearranging as
$$\ddot{\delta} + \frac{1}{RC} \dot{\delta} = \left( \frac{2\pi}{C \Phi_0} \right) (I - I_c \sin(\delta)) $$
we can interpret this as a damped particle with
\begin{align}
\text{"mass"} &= C \\
\text{"coefficient of friction"} &= 1/R \\
\text{"potential energy"} &= -(2\pi / C \Phi_0)(I \delta + I_c \cos(\delta)) \, .
\end{align}
I put the terms in quotes because in the way I've grouped the terms the dimensions don't match up exactly with the mechanical case.
However the roles of the terms are exactly as indicated.
The IV curve
The potential energy term is the key to understanding the IV curve.
Please refer to Figure 2 when reading the following description.

Figure 2: The IV curve and illustration of the junction speudoparticle for a sequence of current biases.
At $I=0$ the junction experiences a cosine potential.
In this case, it sits trapped in one of the wells of the cosine, as indicated in Figure 2 a.
As we turn up the current we introduce the linear term to the potential.
If the bias current is less than $I_c$ there are still minima in the potential and the junction remains trapped as indicated also in Figure 2 a.
Because the junction is trapped at a fixed value of $\delta$, the voltage is zero (recall $V \propto \dot{\delta}$).
This is the part of the IV curve wherein we increase the current without increasing the voltage, as indicated in the horizontal blue line marked $a$ in Figure 2.
Note that at this stage all of the current flows through the tunnel element and none flows through the resistor.
As we increase the bias current past $I_c$, the linear term in the potential overtakes the cosine part and the minima disappear.
The junction then "rolls down hill" as shown in Figure 2 b.
Because we now have a time varying phase, the junction voltage rises from zero to a finite value, as indicated by the red line marked $b$ in Figure 2.
At this stage, $I$ exceeds the critical current of the tunneling element and so the tunneling element no longer behaves as a superconductor.
Quasiparticles are generated, rendering the junction resistive.
Further increases in current show an accompanying linear increase in voltage according to $V = IR$, as shown in Figure 2 c, and by the green line marked $c$.
As we lower the current we travel back down the green line, as indicated by the mark $d$.
The rest of the process depends on how fast we are raising and lowering the current.
As we lower the current below $I_c$, the potential regains its cosine nature and has minima.
If there were no dissipation (or we're sweeping fast enough that whatever dissipation there is can't completely stop the particle during the sweep) the "particle" would continue to "roll" because it already has kinetic energy.
Therefore, even as $I$ is lowered below $I_c$ we still have time varying $\delta$ and therefore still have a measured voltage.
This is shown in Figure 2 e and the magenta line marked $e$.
Finally, when we get back to zero bias, the cosine is flat and as we sweep back toward negative bias we slow down and finally stop the particle.
Then as we increase the negative bias the process starts all over in reverse.
Remarks
If you look closely at the IV curve in the original post you see that it's not a true square as shown in Figure 2 of this answer.
This has to do with the detailed dynamics of the junction during the time dependent sweep.
The damping of the junction and the sweep rate of the current bias play important roles in the shape of the IV curve.
In any case, we have shown that the vertical line at $V=0$ on the oscilloscope trace, which corresponds to the blue line in Figure 2 of this answer, comes exactly during the condition when there is zero voltage but a nonzero tunneling current.
P.S.
The bounty note asks for credible and official sources.
This answer uses Kirchoff's laws and the Josephson relations, both of which are, in my opinion, rather official :-)
A: To me it seems like DanielSanks answer, while being correct, doesn't actually explain the picture at all.
When looking at Josephson Junctions there are two models that can be applied. The first is to drive the JJ using a fixed applied current, I, as explained in DanielSank's answer. That is the Current-Source model. The second model is to apply a fixed constant Voltage, V, known as the Voltage-Source model.
Again starting out with the same equations DanielSank uses:
$$ I = I_c \sin(\delta)$$
$$ V = \frac{\Phi_0}{2\pi} \dot\delta $$
For simplicity we assume that $C=0$. We also ignore the shunt resistor. This is possible because in the Voltage Source model all it will do is add on a constant current $I_n = V/R$ independent of what the JJ is doing. For a constant applied voltage we then find:
$$I_s = I_c\sin(\delta_0 + \frac{2eV}{\hbar}t)$$
i.e. we get a DC current of $I_s = I_c \sin(\delta_0)$ at $V=0$ and an AC current for $V \neq 0$. This is enough to explain the graph. At $V=0$ we get the large current spike seen in the picture. However at $V\ne0$ we get an AC current for which of course $\langle V \rangle = 0$. In general on an I-V graph we are concerned with the DC time-independent currents. This explains why there is no current plotten on the graph for non-zero V. Lastly the step at large V is caused by a break-down of the superconductivity (of sorts).
Which of these two models (Current vs. Voltage source) is appropriate of course depends on the details of the junction and how it is fed. But to me it seems like the picture the OP is asking about most likely refers to a Voltage Source JJ and is better explained using that model. 
A: I think there is still a current flowing. I found this picture

on this website. The gist is that above a certain voltage $V_c=2E_g/e$ (twice the energy gap divided by elementary charge), the voltage large enough to overcome the band gap and Cooper pairs can flow.
EDIT: Also look at this excellent post.
A: To be honest, I can't figure out what does DanielSank show in the potential landscapes in Figure 2 nor the discussion after that.
I think the correct short answer is that the measurement shown in the picture is done effectively by a current bias, not voltage bias. Therefore, the voltage on the junction drops to 0 as soon as supercurrent starts to flow. Think of voltage as a function of current and not the other way around. 
Why does the supercurrent have so many different values (vertical line at ${0}{V}$)? If you current bias, the current sets the phase difference across the junction, see the Josephson's equations in DanielSank's answer. If the current is constant, the phase is constant and there is no voltage drop. 
If you would actually voltage bias the junction, and you would be able to measure the small current through the junction, you should see current oscillations - AC Josephson's effect. 
