The battery charging circuit is in the phone not in the "charger" which is just a dumb 5V power supply.
The issue is that the circuit in the phone will have a maximum input voltage before things break. This circuit will be designed to assume a maximum normal input voltage of around 5.5 V (USB is up to 5.25 and you need to allow a little margin). Your 6 V lead-acid battery will output around 7 V when fully charged.
Is 7 V enough to break things or will the phone cope with it just fine? Impossible to say, it depends on the internal components used and what the designers considered a sufficient safety margin.
Assuming the phone doesn't break it is going to charge slowly, unless the data lines are held in the correct state the phone will assume it's in a standard USB port and not exceed the power ratings of the USB spec. In order to pull 1 A charge current you need to set the data lines to the correct fast charging configuration, this is normally a couple of resistors. I can't remember the exact values needed but a search this site will find the values, it's been asked before. Take care, going significantly over 3.3 V on the USB data lines will break things.