I know that if I connect an electronic component that consumes more than 20 mA, my Arduino Uno will be destroyed. I don't know if that refers only to digital pins or refers to +5V pins too.
If you connect a motor to the +5V pin, it will always be on. Is that what you want?
The +5V pin on the Arduino Uno gets power from the on-board 5V linear voltage regulator OR from the USB jack, and thus the host computer. This allows the board to have a fixed 5V source even when applying 9 or 12 V to the barrel jack or the RAW pin, and to be powered by a USB host.
When you draw 600mA for your motor (ignoring what the other devices on the Arduino draw), you are drawing 600mA from the regulator's output. You are also drawing 600mA from the regulator's input, which supplies a higher voltage. Excess power is dissipated as heat. 5V * 600mA is 3W. If you power the board from 9V, you are discarding [(9V-5V)=4V] * 600mA = 2.4W of power. 2.4 Watts is probably too much for the little regulator to handle. It will get too hot and burn up.
If you do this, you will probably be able to still operate it powered by the USB port, as only the 5V regulator is damaged.
However, if you try to draw 600mA and power the board from a USB device like a computer, you may burn up the computer's USB chip and make it useless forever.
For high-power devices like motors, always power them from a sufficient external supply of the appropriate voltage. Be sure to connect the ground of the external power supply to the GND pin of the Arduino. Don't send high current through the Arduino board. The Arduino should handle signals, not power.
Use the signals to activate switching devices like MOSFETs, which are designed to control higher currents by low-power signals.
No, it shouldn't destroy it. Although you are getting close to the current limit.
The on-board regulator (depending on the board version) can supply up to about 800mA to 1A.
However, the actual maximum current is dependent on the voltage you are supplying the board with - the higher the voltage the hotter the regulator gets and the lower the current that it can supply without shutting down (or burning out).
As long as you run the board from the lower end of the input voltage (7V, say) it should be fine.
It would be better, though, to use an external power supply for the motor anyway, just to reduce the amount of heat the regulator will pump out.
Powering from the USB won't be possible, however: the USB port has a 500mA fuse in the supply circuit.