What is a safe max rated current percentage for a 1A, 5V positive voltage regulator like a 7805?
The maximum current of a regulator is only one limiting factor.  Another, usually more restrictive, limitation, is the total power dissipation.  With a linear regulator like the 7805, the power dissipated by the regulator will be equal to the difference between the input voltage and the output voltage multiplied by the output current.  So if your regulator is providing 5V@660mA from a 9V supply, the regulator is dissipating (9-5)*0.660 = 2.64W.  If you have a regulator in a TO220 package attached to a heatsink, that could be okay, but without a heatsink the regulator will likely get too hot.  

To check, you would need to look up the relevant thermal resistance of the package and multiply this by the dissipated power.  This will give you an estimate of the difference between the internal temperature of the regulator's die and either the ambient temperature or the temperature of the heatsink tab, if your regulator has such a tab.  Add that temperature to the ambient or heatsink temperature, and compare this to the maximum rated die temperature of the IC.  Note that since the die temperature will depend on the ambient temperature, you may find that the die temperature will exceed ratings in some conditions, but not others.  So it's important to know the environmental conditions in which your system will operate.  

Also note that when a heatsink is involved, its thermal resistance will come into play as well.  If you know the approximate thermal resistance of your heatsink in your expected operating conditions, then you can add the heatsink's thermal resistance to the regulator package's thermal resistance, and obtain a total thermal resistance between the die and the ambient air.  However, the performance of a heatsink is highly dependent on airflow conditions, so in some cases the heatsink's characteristics must be measured in situ (or carefully simulated) for accurate results.