When dealing with leakage currents you need a much higher impedant set-up. Remember, leakage currents are very small and can only be detected by their voltage drops aross large shunt resistances. Putting the diode in parallel to a 300R resistor, as in your set-up, is a way too weak "detector".

To measure the leakage current of a diode by the help of DVM (digital volt meter), connect the anode of diode to the plus pole of DVM. Now connect a positive supply voltage of 20V with its plus pole to the cathode and with the minus pole to the minus pole of DVM. Now, a leakage current will flow through the diode and the internal input impedance of DVM, which is about 10MOhm and serves as high ohmic shunt resistance. The DVM displays the voltage drop across this impedance, which is

U = I[leakage] x 10MOhm.

From the reading you can find the leakage current.

Example:
Reading shows 0.5V. Then, leakage current is 50nA. Test voltage is 20V - 0.5V = 19.5V, means about 20V.

For an embedded set-up I would recommend a chopping scheme, where the test voltage is a square wave, toggling from 0V to 20V. The rest to do is to amplify the voltage drop across shunt resistor, have the signal running through a high pass filter to suppress the DC component and to finally rectify the AC component.
The chopping scheme makes your methode first order immune against input bias currents of OPamps, parasitic input leakage currents resulting from eventual protection circuits at the input of first amplifier, their voltage drops, offset voltages, parasitic thermoelectric potentials and all kind of offset drifts.

Kai