Why is a Josephson junction important for a SQUID? A SQUID works using the concept that flux going through a superconducting circuit is quantized. So, if we try to change the magnetic field smoothly, the circuit will produce a screening current that would create a flux that cancels the difference between the applied flux and the previous/next quantum of the flux depending on how close we are to each. The measurement of the screening current tells us how far we are from a quantum of flux, and from that we may sense the changes in a magnetic field applied to the circuit.
In this I described how a SQUID works and I didn't need any concept that requires the Josephson junction. What's the use of a Josephson junction in a SQUID and what would happen if we build a SQUID without a Josephson junction?
 A: You confound screening current and quantisation of flux. They are not the same. Screening currents come from the perfect diamagnetism of the superconductors, whereas the quantisation fo flux come from resistless currents. They are both due to the uniqueness of the macroscopic wave function, understood as a charged macroscopic particle, especially the phase: in short, if you can write the wave-function of the superconductors as $\Psi=\Psi_{0}e^{\mathbf{i}\phi\left(x\right)}$ with a non vanishing $\Psi_{0}$, and if the free energy of the system reads 
$$F\propto\left|\left(\nabla-\mathbf{i}\dfrac{A}{\Phi_{0}}\right)\Psi\right|$$ 
($\Phi_{0}=\hbar/2e$ the (superconducting) quantum of flux, $e$ the electronic charge, $A$ the vector potential and $\mathbf{i}^{2}=-1$) then you have both quantisation of flux and screening current. I believe that's the reason why you confound both.
Now, imagine a piece of superconductor (say, it's a rectangle) with a hole (say, it's a circle) inside. You have screening currents along the rectangle and the circle for sure. Let's go to the quantisation of flux. It states that the flux inside the circle is quantised. And so what? What does it has to do with the screening current you have along the rectangle, along the circle? More importantly: which one (circle or rectangle) could you hope to measure with an amp-meter? 
So let us turn to the JJ: you there have the fantastic relation 
$$I=I_{c}\sin\phi,$$
which tells you that if you put an insulating barrier between two superconductors, you can transcribe the phase-difference $\phi$ between the two-superconductors to a current flowing through the junction. So we go back to our previous system, and we shrink the rectangle down to touching the circle at two points, and connect the rectangle to any amp-meter. Then we have a superconductor, a hole inside, two JJ along a diameter of the circle, whereas an amp-meter is contacter perpendicularly to this diameter (it's called a dc-SQUID if you want to look for pictures on the web). Since the phase-difference records the flux inside the circle, so will do the current passing through the system. A dc-SQUID has actually 2-JJ such that the measured current is $\propto \cos \pi\Phi/\Phi_{0}$, the flux inside the loop being $\Phi$.
I hope the above arguments convinced you that JJ are indeed important to record the flux: the Josephson physics around a closed loop transduce the flux to current around the loop, what the screening currents do not.
