MOSFET load switch design review
The control voltage is 3.3V

The main problem
The gate activation voltage for Q7 is only 3.3 volts!
It will not turn-off the PMOSFET inside Q7 and the circuit will be permanently energized.
Or, ditch the dual MOSFET device (Q7) and use a BJT in common emitter with a load resistor of circa 1 kohm (aka a level shifter). General idea: -

This can only realistically be overcome with either a control voltage that matches the supply voltage (9 volts to 12.6 volts) or by using an extra input transistor forming a level shifter. This is a show-stopper until fixed. The words below assume it has been fixed.
Given this: -

There is no strict requirement on the switching frequency (less than
1Hz)

Using a level shifter isn't going to be a problem.
The secondary problem
Always check with the manufacturer's website to see if the a device is not recommended for future designs and if so, is an alternative available: -

However, given that this device is probably best replaced with the common emitter level shifter shown earlier, this is moot.
Shoot-through problem
This can be overcome by using a 100 Ω resistor in series with the drains on Q7. The shoot-through current will be limited to a maximum of 12.6 volts ÷ 100 Ω = 126 mA. If that is too much choose a 1 kΩ resistor etc.. The added resistance will slightly slow down turn on or off but, given that you are only switching at 1 Hz and the gate capacitance of Q2 is only circa 3 nF, a response time of less than 5 μs isn't going to be a worry.
But if you fix the main problem using a BJT level shifter, this problem goes away.
Load volt-drop

If you can live with Q2's volt-drop of 40 mV when the load takes 10 amps then you're basically good to go. That's a typical value and may be twice this amount across a batch of devices.
MOSFET leakage (\$I_{DSS}\$) into load
When Q2 is off, it might produce typically 1 μA into the load but, again, this is a typical value and it might be as high as 100 μA. If this isn't going to cause activation of some low power circuits in the load then fine.