Is this SHA256 hash implementation secure from rainbow table, brute forcing attacks?
TL;DR: 64-bit entropy is nowhere near enough against a serious attacker.

Yes, you have calculated the worst case correctly.
Your second calculation isn't about rainbow tables, but about a full hash table.

Now for "how long can you stop a serious attacker" if you hash a fully random 8 byte string with SHA-256. If the attacker has enough money, they can design their own ASICs for this or ask nicely at companies like AntMiner to produce such ASICs. So assuming you pay a little more, you can probably get 1GHash/s for 1000 USD (AntMiner's S7 is at 4.7THash/s for 575USD + energy, network,... at the moment). So to break the hash using one miner (with 1THash/s), you need 213 days, whereas if you buy 427 of them, you can break the hash in at most half a day and expect to break it in a quarter of a day on average and at this point you wouldn't even be egligible to the highest discount they offer (at 500 pieces and more).

Rainbow tables should allow for quick breaks of the security and normally reduce the storage required to $2^{2n/3}$ and the computational requirements to $2^{2n/3}$ per look-up or more generally, given $2^m$ storage, the lookup time will be around $2^{2(n-m)}$.
If you want to defend against rainbow tables, salts are inevitable, because you need a full rainbow table per unique salt, which is computationally and storage-wise intense. As for the efficiency of rainbow tables, you need a full pass over the complete space (all $2^{64}$ values) to build the table. For the original (first) publication of rainbow tables, see "Making a Faster Cryptanalytic Time-Memory Trade-Off" by Oechslin (PDF)