Well, that's the recommended circuit. Keep in mind that at the 2uA current draw of the RTC backup, even a 1N4148 will have only about 0.3V drop and it will work down to Vrtc = 2V.
A CR2032 lithium primary cell depleted to 2.3V at 2uA is practically dead, so you're not leaving many (micro)joules on the table.
Edit: In explanation of the final paragraph above, let's refer to the Panasonic datasheet for their CR2032 battery:
This does not directly answer the question of what the discharge looks like at 2uA, but at 190uA you can see that the voltage drops fairly abruptly and a 'cutoff' voltage of 2V vs. 2.3V buys you minimal additional hours. Of course you would expect the times to be about 100x longer for 2uA vs. 190uA but the principle is similar. You're extracting almost all the energy (as measured in joules) from the battery if you consider end-of-life to be 2.3V and setting it at 2V or 1.5V buys you very little additional time. So the diode drop of 300mV nominal is really not very important to the total battery life on standby.