3.3V Buck-boost converter outputting voltage too high
Your circuit will likely work with an inductor with an appropriate current rating.  Remember that the load current is the average of the current in the inductor.  The current has a triangle waveform; in boost mode, the inductor is switched to ground to "charge up" the inductor until its current is greater than that required by the load.  Then the switch is opened and the inductor is current is dumped into the load, with the current decreasing as the inductor "discharges:"



In buck mode, The voltage input voltage is higher than the output and the inductor charges up while the supply is connected to the load through the inductor.  The supply is then disconnected and the inductor current must flow through the diode.  Since the inductor has a negative voltage, the current decreases:



The inductor B_H Curve looks like this:



The operating range of the inductor must be in the relatively linear area.  Since the inductor only has current flowing in one direction, it must operate in the upper right quadrant only.  The area marked "minor B-H loop" would be acceptable, with the BH loop sliding up and down the curve as the inductor charges and discharges.  However, too much current will drive the operating region up into the non-linear part of the curve.  The inductance is not constant in this area and it will drop and change value at different points in the boost or buck cycle.  

In boost mode, the inductor must store enough energy on each charge cycle (1/2Li^2) to provide average power on the discharge cycle.  You just don't have enough core to do this when L drops due to saturation, and the result is too low voltage.

In buck mode, the inductance drops during the charge cycle, causing a current surge into your load and capacitor and fully charging your inductor.  The load must then discharge the inductor during the time before the next charge cycle.