Digging even deeper you will see that there are two figures in the PIC datasheets which describe the output voltage level (VOH) as a function of loading (IOH, in mA), each with graphs for three different temperature levels. One figure is for Vcc = 3V, the other is for Vcc = 5V. These figures reveal that the 0.7V voltage drop (resembling the goodol' bipolar transistor base-emitter voltage drop) mentioned in the table is rather conservative and may be relaxed in practice when the loading is only a few milliamps. The figures are identical for the different pic types. See also figures 18-28 and 18-29 in the PIC16F886 datasheet (28X1).
In practice this means that for a loading of 1 mA the voltage drop is approx 0.2V (please have a look at the figures; they are not completely linear, and differ slightly for different Vcc's). Since many interesting ICs work in the few mA to less than 1 mA range, the voltage drop will be neglegible in many apps.
For larger loadings this non-optimal PIC output behaviour becomes apparent, but can be turned to an advantage in LED applications when the Vcc supply voltage is 3V (not 5V !). This is because a LED has an almost constant forward voltage drop of 1.6-2V with varying current (in other words, the LED brightness depends on the current, not the voltage over it), the actual drop depending on the LED type and color. This means that often for 3V apps the current limiting resistor can be dispensed with; coupling the LED directly with the output causes the current to settle somewhere between 5 - 15 mA (and the voltage at the LED forward drop), which is in the operating range for both PIC and (most) LEDs. I suspect that this is the reason that I have seen some PIC apps at 3V without LED current limiting resistors. Do look at the LED spec (or simply measure the voltage drop yourself).
/Jurjen
