Is there an easy way to tell whether this "Curlmeter" would rotate or not? Let us say we have the following symmetrical apparatus:
 
Four equal positive charges, all connected to a shaft that can rotate, the connecting rods are insulated, and so does the shaft.
Now suppose that a positive charge, less in magnitude than the four previous ones, pops out in the indicated location below:

Is there any way to tell whether the apparatus would spin without brute forcing the problem with Newtonian mechanics?
 A: In the example you drew, there will obviously be a clockwise torque, as you indicated with the arrow. You can make the extra charge as close as you like to one of the arms of the curl-meter, so that force can be as large as desired.
The curl-meter is supposed to mock up the curl operator, which is a local thing. Derivatives are local. That means that you need to imagine taking the limit as the curl-meter's size approaches zero.
The OP asks in a comment:

The confusing thing is since the curl of an E-field is null, then so must be its circulation around any loop, so the work done around the circular path by the force generated by the small charge must be zero, and so the curlimter shoud not turn at all, or is it that it rotates clockwise and then gets back to its original position, i.e., it oscillates? 

If you rotate the large meter (which isn't really a curl-meter because it's not infinitesimal in size), then the work done is $\int_{\theta_1}^{\theta_2} \tau d\theta$. This work will be zero if you integrate over one full rotation, because this field is conservative. It doesn't matter whether you consider free rotation or not. The work done by the electric field is zero, for example, if you rotate the meter by hand using a crank. The fact that the work integrates to zero over one cycle does not imply that $\tau=0$ for a particular value of $\theta$.
