Disabling IC by removing VCC
If your intention is to make sure that the voltage rails are nearly zero when enable is low, you'd have to drain the voltage store at these capacitive nodes, either passively or actively. 

Passive Draining

Draining it passively is accomplished by connecting a resistor in parallel to the voltage rail of interest. A low resistor will lead to a continuous power loss in the system during normal operation, but will make sure to drain the node very rapidely once you disable the optocopler. For a highly resistive resistor, you would have to wait for some time until the supply voltage rails would go down to an acceptable level.

Since you are using \$15V\$ and \$5V\$ in your system, bleeder resistors  would have to be connected to both rails. This approach is the easiest one and require small modifications, however, it has the drawback of adding additional power loss.

Active Draining

Implementing an active draining circuit, reduces the continuous power loss but it comes with the burden of adding more complexity. The idea here is that the nodes are only drained once you disable the optocopler. For the the \$15V\$ rail it would look like something like:

Here, the \$V(out)\$ which represents your \$+15V\$, is drained through a \$100\Omega\$ resistor and a pair of MOSFETs. The draining "speed" can be configured through the active bleeder resistor \$R_3\$.