Why don't you just use more of the one inch square magnets? Let's take a look at this. I assume the design calls for nine coils/twelve magnets. So, how to begin? Let's start with the amount of copper (resistance). The total amount of copper per coil is proportional to the area of copper wire. I will also assume that the total winding thickness is one half inch. So, for the 2 by 1 inch magnets one gets:
3x2 - 2x1 = 4 square inches of copper.
For the 1 inch by 1 inch magnets one gets: 2x2 - 1x1 = 3 square inches of copper. So, you can increase the number of coils by one third and still have the same resistance. So, go to 12 coils/16 magnets of the same thickness and the same number of turns of the same wire. With the greater number of coils we now need a larger rotor, one third larger, in fact. The larger rotor means the magnets pass the coils faster for the same rpm. We have increased the total flux by one third and the time rate of change of the flux by one third. To get the change in voltage relative to using the one inch square magnets in the 9/12 configuration we multiply 4/3 times 4/3 to get 16/9, about an 80% increase in voltage with the same resistance. This isn't too bad, as going to the 2 by 1 inch magnets isn't going to double the voltage anyway due to geometric limitations. When the coils are centered over the magnets so that the flux is greatest, the voltage waveform from a coil is actually going through zero. The voltage maximum occurs when a coil is centered exactly between two magnets. Unfortunately, the copper wires will not be exactly parallel to the magnet sides, which means the voltage will be somewhat lower than optimum. Also, the part of the 2 inch magnet closer to the center of the rotor will be moving slower than the top half, which also acts to lower the voltage.
If you are still concerned about having the proper cut-in rpm you could sacrifice resistance by increasing the number of windings by 5% and the rotor diameter by 5%, for a 10% increase in voltage.