USB C Powering over VCONN with no VBUS
As I understand, you mean that your host has a 5V internal battery supply that can afford only 200 mA to source (VCONN is 5V @ 1W), right? 

The USB PD protocol indeed is the correct thing to advertise "Battery Supply Power Data Object" (Table 6-9 of PD specs) with B[9..0] = 0x004. The use of VCONN is not defined/allowed to power any slave USB devices, only e-markers and PD chips, and VCONN is not connected to VBUS anywhere. 

However, based on declared backward compatibility with USB 2.0, you can declare your host as "low-power embedded host", see section 7.2.1 of USB 2.0 specifications (or Section 11.4.1 in USB 3.0), "Classes of Devices", which says, in "Root port hubs" sub-section,


  "Battery-powered systems may supply either one or five unit loads."


So, one unit is legal. If you declare your port as "low-power" (in a formal USB-IF declaration for certification), you can legally supply VBUS with only one unit load capability, 100mA for USB 2.0 protocol, and 150mA for USB 3.0 protocol. 

However, there is a gap in USB Type-C specifications. While the Type-C specifications explicitly refer to USB 2.0 standard with regard to "default power", the Type-C connector does not define a special pull-up value to advertise the "low-power" port capability. So you can use only the standard CC pull-up of 56k, for "default power". Here is the rare place where the USB device restrictions come to play, to draw only one unit until enumerated. 

To make it work without blowing up your host, your host software driver must deny any device that specifies its power consumption above your port capacity (200mA in your case), and hardware must include an overcurrent protection of your port at 200mA as well, to prevent badly-designed USB devices from drawing too much current without being enumerated/configured.