Unfortunately, yes, if the USB 2.0 device seeks the USB-IF certification, it must obey the sequence 100mA -> USB_connect -> enumerated/configured -> full draw of port current. The "full draw" however should be determined by sensing the value of pull-up on attached CC pin.
If the attached cable is a "legacy cable" compliant to Type-C specifications, it should have 56k built-in pullup (you generally don't need any adapter, the cable has the pull-up).
If the attached cable is the "Type-C Standard Cable" connected to another Type-C port, the CC pull-up will be defined by the sourcing port, whichever it is capable of, 1.5 or 3A.
If you want your device to be practically charged, and charged faster, it is advisable to implement a battery-charging detector IC, at least to determine if the port supports Chinese-style charger signature, D+ connected to D-.
To really comply with USB-IF test specifications when using Type-C connector, you need to consult very carefully with this document, Type_C_Functional_Test_Specifications. This is an evolving area, so check for updates.
But if you don't bother with exhaustive USB-IF certification (as most manufacturers do), just take 500mA if it is enough for you, since every reputable host USB port must unconditionally support 500mA of sourcing (except nearly non-existing subset of low-powered portable gadgets running form tiny batteries).
The 28AWG wire has resistance of 65 mOhms/ft, see Wikipedia. With 7 feet two ways (ground and VBUS, = 14 feet) the total parasitic resistance added by the wire is about 1 Ohm, if account for connector contact resistance.
An external hard drive can take 800-900 mA of current, some up to 1.8 A when spinning up, which would degrade the input voltage by about 1 V (1A over 1 Ohm). This drop would be out of specs and is usually devastating for mechanical hard drives.
So yes, in this particular case the weak spot is power delivery over 28AWG wires. The next hurdle would be the impedance (mis)matching across the cable data wires and soldered connectors.
Best Answer
USB-PD only uses two wires (GND and VBUS), but the standard has the whole section (4.4 in 'USB_PD_R2_0 V1.0 - 20140807.pdf') which describes how the cable detection is implemented. In short, regular USB-A to USB-B cables need to have the special mechanical mark to be considered for >5V/>1.5A, while micro-usb send special signal on "ID" pin, and determines capabilities from that.
In theory, if you do not use special PD cable, then the devices will silently limit the maximum power to pre-PD levels, and all will be safe (at most 5V, 1.5A current).
In practice, I bet there will be manufacturers which produce very cheap cables which still claim to support maximum power, so we might still read about cheap cables catching fire.