Would SHA1 be broken by sheer brute force, even if it had no weaknesses of its own?
Yes, it is now conceivable to find a SHA-1 collision by brute force.

The general method is outlined in Paul C. van Oorschot and Michael J. Wiener's Parallel Collision Search with Cryptanalytic Applications (in Journal of Cryptology, 1999; free earlier version available from the first author's website; expands on an article presented at CACM 1994). It uses local search iterating hashing; when a distinguished value (say with a certain number of zeroes in low-order bits) is found, it is reported to a central location; enough findings lead to a collision. It is expected that less than $2^{82}$ hashes (and comparatively little extra work) are required to find a collision.

By May 2018, the bitcoin mining aggregate hash rate is reportedly $30\cdot10^{18}$ SHA-256d per second ($2^{81.1}$ per day), where SHA-256d uses two SHA-256, and SHA-256 is roughly as costly as SHA-1. The ASICs predominantly used for that can not be directly repurposed for a SHA-1 collision search: they use SHA-256d, not SHA-1; and as an aside they hash incremental values, rather than iterated ones. However the technology is there.

Thus the energy spent in a day of bitcoin mining is enough to find a SHA-1 collision by brute force, hypothesizing the same technology is used. The collision found would be a random one, with chosen prefix decided at start of attack, and could not be detected.

And it might well be that significantly less work would be enough to find a collision that evades detection using the current "fingerprint" method, by using specially crafted (but less efficient) differentials than used to find the first collision reported.