For low power DC-DC converters the quiescent currents of the parts involved are usually what hurts the efficiency. Also, the switching losses become a problem, which is why you want a low frequency operation or other techniques, which, in fact, produce an equivalent low frequency.
Personally I would build the boost using a real power supply controller, which offers other features, such as current limit.
Having said that, I will go back to your question.
The issue with the switching losses can be improved by switching the MOSFET faster. Many people misunderstand the PIC's "high current capability". That is just a measure of the internal channel resistance of the transistors. Hence 5V/25mA=200ohm. That is it. So basically you are driving the MOSFET with a gate resistance of about 200ohms (PIC running at 5V). With these in mind we’ll do an analysis of the losses in your converter.
The switching time of the MOSFET is best calculated from the gate charge graph, a picture that looks like the one shown in the first half of my drawing.
The area below the horizontal "plateau" in the middle is called the Ogd. The MOSFET switches on when that charge has been transferred to its gate. But the charge is transferred through the gate resistor and, as you can see, the gate voltage is not 0V, but Vth. Therefore, the current you actually supply to the gate is (Vmax-Vth)/R. So it's not 25mA, as you might think, but much less. Since your PIC is runnig at 5V, Vmax is 5V. Vth is about 3.5V for your MOSFET. That value has to be adjusted for your actual MOSFET current. Vth will be closer to 2.5V at your operating current, and switching times about equal.
So the gate current is:
Ig=(5-2.5)/200=12.5mA
Now, Qgd is from the table 9.3nC max. So your maximum switching time will be:
tsw=Qgd/Ig=0.75us
This is a fairly long time, but assuming a switching frequency of 15.625kHz (T=64us), your switching losses will be (assuming CCM operation):
Psw=Vout*Iout*tsw/T=2.6 mW.
This is about 1.1% of your total output power.
There will be losses associated with driving the gate, given by the total gate charge, Qg:
Pgate=Qg*Vmax/T=1.17mW (Vmax is 5V, since PIC is running at 5V)
Duty cycle at min input voltage and max output power is:
D=1-Vin/(Vout+Vdiode)=0.675
The MOSFET RMS current will be approximately (under CCM conditions):
Irms=Iout*sqrt(D)=12.3mA
MOSFET conduction losses, approximately (under CCM conditions):
Pcond=Rdson*Irms^2=0.009mW very low, which is normal for this low power converter.
Losses associated with Coss (250pF at 15V):
Pc=0.5*Coss*Vout^2/T=0.44mW
Diode conduction losses (under CCM conditions, max output power/ min input voltage):
Pdiode=Vdiode*Iout*(1-D)=1.95mW .It will get worse at higher input voltages, because D decreases. However, I assumed that at maximum output power your input voltage will always be at its minimum, since you are running from a solar panel. Vdiode is about 0.4V.
There will be losses associated with the inductor. I will assume a resistance of 3 ohms for the inductor, since for CCM you need a large inductance, which means higher resistance, too.
Pind=Rind*Iout^2=0.67mW
For the PIC, the quiescent current could be about 2mA @5V, that is Ppic=10mW. But if the PIC is running from a linear regulator you have to consider those losses, too.
So, now we have: Ploss=Psw+Pcond+Pgate+Pc+Pind+Pdiode+Ppic=18.04mW
Efficiency is Eta=Pout/(Pout+Ploss)=92.6% (assuming no other loss for PIC regulator)
As you can see, you can only reduce the losses associated with the MOSFET, by using, for example, the NTR4503 from ON Semi or the TN0201K from Vishay (Siliconix), which have much lower gate charges and Coss. Or you can use a gate driver, like the 2 transistors, which allow you a much lower resistor, say 33 ohm. But the big player remains the PIC quiescent current. To reduce diode losses you would need a synchronous rectifier, but that one will also have gate drive losses, so the gains may not be too great.
As for your approach, you will have to use continuous current mode (CCM), because that will give you better regulation, although not good regulation. With discontinuous current your output voltage would depend too heavily on the output current. The best thing is to sense the output voltage and change the duty cycle.
In either case, your regulator has no MOSFET protection.
There can be a change in efficiency as the input voltage changes, but it will always be less than 100%. Physics...