USB 2.0 high-speed on all ports

This is a misconception. Nobody "splits" or "re-distributes" anything in USB3 architecture. USB2 signals go through USB2 D+/D- channel exclusively, USB3 signals go through Tx/Rx pairs. I guess this misconception comes from the older and different USB2 architecture, where full and low-speed (FS and LS) transactions can be really "split" between multiple FS/LS connections. In USB2, this function was implemented via so-called "Transaction Translator" (TT) block built into each hub. And the HS bandwidth could be "split" between FS devices only if multiple TT (MTT) are designed into the hub. This kind of translation architecture is not implemented nor specified in Super-Speed USB3 architecture.

If you connect four USB2 HDD to a hub, all FOUR will get an aggregate of 35-40MB/s per HOST, or 8-10MB/s sustainable throughput per each drive. Higher USB3 "bandwidth" has nothing to do with USB2 transactions, absolutely nothing (except that the xHCI USB3.0 controller architecture might be somewhat more efficient in handling protocol overhead).

USB3.0 hubs have essentially two hubs inside, a USB3, and a legacy USB2. They function independently, they just redirect (mux) downstream ports to one hub controller or another if a device is attached, or just work in parallel if hubs are in the segment.

Same goes for USB3.0 hosts, the architecture is the same, and they are called the same, except that it is "root" hub. If you get the USBview.exe utility, you will see that the root xHCI controller branches into TWO ROOT HUBS, one SS USB3, and another USB2. If you plug a USB3.0 device into mainboard USB port, it goes into the USB3 section, and logical port number will be, say "1". If you plug a USB2 device into the same physical port, it will be directed to USB 2 branch, and its logical port will be something like "5", so the total bandwidth will be defined by corresponding root hub section. No one shares the super-speed bandwidth with high-speed devices, they are nearly independent.

So the answer to OP question is "NO".