R6 is too small, it should be in the order of 1M or 470k.
R3 is too large, make it about 4k7. This resistor forms a potential divider with the integrator's input resistor. When the input to the integrator goes high these two 100k resistors are acting together to halve the input voltage to the integrator compared with when the integrator's input goes low which unbalances the integrator.
There are 3 factors which control the amplitude of the output triangle wave:-
Amplitude of square wave input.
Frequency of input square wave. That is to say how long the square wave stays in the high state and how long it stays in the low state.
The R4 * C1 time constant.
At the moment, the amplitude of the triangle wave is far too large (the op amp is saturating at the supply rails) so increase the frequency of the sine wave or increase the size of C1 to bring the triangle wave's amplitude down to within the supply rails.
Even with a change to R3, making it 4k7, the square positive going excursion will be slightly smaller than its negative going excursion causing an offset of the triangle wave from centre ground. To significantly reduce this offset insert a small resistor (say 100R) between the 4k7 resistor and the output of the LM311 and connect the input to the integrator between these two resistors. This will reduce the amplitude of the negative going swing of the square wave and the triangle wave will be more central. You will have to adjust the value of this extra resistor to find its optimum value in order to minimise the triangle wave offset.
You may have realised from the 3 amplitude controlling factors listed above that the amplitude of the triangle wave will vary with sine wave frequency. Fixed value components will only allow a limited frequency range for an amplitude of the triangle wave between very small and an amplitude which is limited by the supply rails.
To calculate the amplitude of the triangle wave:-
Vout = -(Vin * t)/(RC)
Vout = the amplitude of the triangle wave
Vin = the amplitude of the square wave
t = half the time that the square wave is either high or low = 1/(4f)
RC = input resistance multiplied by feedback capacitance
There is a negative sign in the formula because when the input square wave goes positive, the integrator's output ramps down in a negative going direction.