no, what that means is that if you tie the pin to vcc or ground, the leakage current will be 0.3uA max (could be less). This is usually ignored, however (0.3uA ain't much). If you tie it to VCC, the leakage will be +0.3uA (into the device), and if you tie it to ground, it will be -0.3uA (out of the device). This is used when you drive the pins with a gate, you have to know how much current the input leaks so you can design your driver correctly. If you have a driver that only drives 10uA (what? they typically source/sink 10-20mA) and try to tie 40 devices together (40 x 0.3uA = 12uA), you won't be able to drive them all the way to the rails. That is why you see a limit of 10 TTL loads on a TTL driver, because of the input leakage vs output driver capability.
You may also need this number if you are calculating how much power your system needs. Like I said, 0.3uA ain't much, so it's usually ignored. Takes a lot of them to add up to even 1mA. However, if you are using a voltage reference device that sources a very low amount of current, you need to add your leakages accordingly.
It's like on opamp designs, everyone assumes the current in the feedback resistor will be the same as through the input resistor, but it will differ by the pin leakage current (see the opamp spec sheet). But it is usually so low compared to the feedback current, you just ignore it (but know it's there), especially if you're using 1% resistors, your output will change more by resistor difference than leakage current difference.