For any given load, a switcher will transfer a given amount of energy thousands of times per second. This is how the buck regulator works.
Let's say your op-amp is switching at 10kHz (because it's a slow sort of device and will have slew rate problems compared to other devices). Let's also say you are aiming to deliver 5V across a 10 ohm resistor. Resistor power is 25/10 watts = 2.5 watts.
To calculate energy per switching cycle divide this power by frequency because power = joules per second. At 10kHz, the energy you transfer per switch cycle is 250\$\mu J\$.
This energy powers your load resistor but, if you removed your load resistor, this energy gets dumped into the output capacitor and its voltage rises a little (or a lot) higher than normal.
Let's say your output capacitor is 10uF - if suddenly it was imbibed with 250\$\mu J\$, how much would it rise in voltage?
We know that capacitor energy is \$\dfrac{C V^2}{2}\$ therefore we can calculate the voltage rise and this is: -
\$\sqrt{\dfrac{250\times 10^{-6} \times 2}{10\times 10^{-6}}}\$ = 7.07V.
It's a little bit subtler than this - in the above I assumed the capacitor was being charged with energy from a zero voltage state. In fact it already has 5V across it and this means that the previously stored energy + influx energy (from the inductor) is 125\$\mu J\$ + 250\$\mu J\$ = 375\$\mu J\$.
If you do the reverse math, the peak voltage on the capacitor becomes 8.66V i.e. 3.66 volts higher than the 5V rail.
You could put an argument together to consider the losses in the diode also - this may trim half a volt of the absolute peak voltage.
So, you either need to increase the capacitance a lot or, decrease the transfer energy by increasing the operating frequency. Modern switchers regularly operate at 500kHz and this means the energy per cycle reduces from 250\$\mu J\$ to 5\$\mu J\$ in this example.
Should this be the case (500kHz operation), the rogue energy from the inductor would make the capacitor's stored energy 130\$\mu J\$ and this means a peak voltage of 5.1 volts - probably quite acceptable for load dumping on a switcher.
Operating at higher frequencies requires faster silicon but, the ability to control load variations (and their repercussions), on a cyclic basis, means much tighter control of the output voltage.
This is just an example to see where you might be going wrong.
While you may be able to get a 12V@15A buck-boost supply to work, the boost portion of that power supply is the portion of the design that will be very challenging.
Linear Tech does offer the LTC7812 which would supposedly allow you to do this (bear in mind the link is a press release!). You would need to cascade multiple units together to accomplish your goal, and that solution is designed to cascade outputs together. I have not personally used this part before so I can't vouch for the viability of that part.
Linear Tech does have a good reputation, and they probably have some development kits that you can obtain from them that would allow you to play around and see if it works. Once you are past that stage, it is likely to be fairly expensive to build this from the ground up; so unless you are building a lot of units you may wish to consider a solution that is composed of off-the-shelf components.
Have you considered using 24V automotive (truck) batteries and 24V-capable battery chargers? All of these components including the 24V->12V buck converters are readily available and can be sourced from retail outlets like Amazon.com.
I love the idea of rack-mounting this. Don't forget to add some cooling fans as dissipating ~30W of heat due to loss in there is something that you will need to think about. You'll want to ensure that you have some headroom in your battery capacity if you are gone for a long time, especially if the system is going to experience environmental temperature variations as that does sometimes affect battery capacity. This varies from battery-to-battery chemistry, so do your homework on that one before you build a bunch of units.
Good luck!
Best Answer
You could potentially put in parallel several Buck circuits to create the current you need, but I don't suggest it unless the controller has a Sync pin to synchronize current draw.
To be honest, it might actually be cheaper to build a custom synchronous buck circuit driven by a tiny micro controller. At 30A a synchronous buck design is a must, multi-phase might also be a good idea, though that makes it much more difficult to design. I know that's not the answer you wanted to hear, but your power requirements are too specific for a company to have invested in an ASIC for that.