Capacitance on the output of an op amp
Most OAs have limited capacitive load capabilities. The origin is quite intuitive. Consider the simple circuit below. On the left there is a real OA. On the right, the OA has the same characteristics (poles, etc.) but the output resistance is externally shown.

This will have the effect of creating a pole, with time constant \$R_{out}\cdot C\$.

Such pole will reduce your phase margin of your system.

Limiting case: C is so large that it can be considered a short. Then you're actually removing the feedback, and you get a comparator (assuming non zero feedback Z)! 

Instead, if there is a "large enough" resistor in series to C (R19 in your circuit), the series R-C will add a pole (with time constant \$(R_{out}+R_{19})\cdot C\$ ) and a zero (with time constant \$R_{19}\cdot C\$ ).
If, as said, R19 is large enough, the pole and zero time constants are close enough and the overall effect is negligible (limiting case: R19 is infinity).

Intuitively, after the zero frequency, the series R19-C will act as a load, and it will not open the feedback (it will change the feedback attenuation factor, but if \$R_{19}&gt;&gt;R_{out}\$ then this variation will be negligible as well).

EDIT:
In your case, the circuit will be stable. 

EDIT2: the following consideration is no more valid with the new edits to the schematics made by the OP:

I assume that "Input Signal" in your schematics is a current signal (i.e. not a connected voltage source), otherwise the circuit won't work.







simulate this circuit &ndash; Schematic created using CircuitLab