Hello....
I have the following circuit that I have breadboarded and simulated. When turning the 60V power on while R1 disconnected the load power comes up. Even with R1 connected as shown and cntrl kept low (without driver) load power comes up when 60V is powered. Can someone please tell me how to solve this? Is this leakage current that is creating the grief?

 

What is the purpose of M2? You can never turn it off because the body diode is forward biased.

 

What is the purpose of M2? You can never turn it off because the body diode is forward biased.

The title refers to a bidirectional switch and if the 60V supply polarity was reversed then M2 comes into play.

One problem I see with this circuit is that once C1 is charged there is no way to discharge it again. Once the switch is on, there is no way to turn it off.

 

Use an Isolated-Gate-Driver and 2 N-Chan-FETs .........
.
.
.

.

 

Post your .asc file.

One problem I see with this circuit is that once C1 is charged there is no way to discharge it again.

Yes.
Try putting a 1 megohm resistor across D1 and reduce the value of R1.
(What is the purpose of C1?)