Not really. The standard claim that if you go to high enough energies string theory predicts that you will see string excitations with linearly rising energy and characteristic string theory scattering amplitude behavior are based on perturbative string theory. Non-perturbatively, it is very unclear if this behavior will survive: what if you start producing black-hole states, for instance? The standard conjecture about perturbative string theory is that it is just one special corner of a theory called “M-theory”, which generically does not necessarily contain string-like states at all. Yes, if you were to see characteristic string-like behavior at the Planck scale, you would have good evidence for string theory, but string theory unification does not require this at all, with many models not having this behavior.
If you don’t believe me, maybe you’ll believe Arkani-Hamed, see this posting, which includes:
In the question session, he made the same point I often end up arguing with string theory proponents about, saying (1:14) that if “you can do experiments at the string scale, wouldn’t help you at all”. The idea that you would see string excitations on a compactified space he characterizes as a misguided old idea from the 1990s. If there’s a landscape, the possibilities are so complex for Planck scale behavior that you can’t predict what experiments at that scale would see.
For more, see here
Update (11/2018): For a related argument that string perturbation theory doesn’t sensibly predict anything at the Planck scale, see this preprint from Banks and Fischler:
String theorists have avoided thinking about this problem because the perturbative S matrix has finite matrix elements in Fock space, once one goes to sufficiently high dimension. However, as the four dimensional case shows, the real issue has to do with infinite numbers of arbitrarily soft gravitons. This is related then to the behavior of the perturbation series for very high orders, and we know that it diverges badly[17]. Indeed, when one thinks about the physics this S-matrix is supposed to describe, it becomes obvious that no perturbative treatment of this question is adequate. For example, it is widely believed[18] that scattering of two gravitons, at an impact parameter smaller than the Schwarzschild radius of the center of mass energy, will produce a black hole. The gravitational S-matrix in any number of dimensions, thus describes processes in which scattering of a finite number of particles produces a collection of large black holes, which can orbit around each other emitting gravitational bremstrahlung, coalesce, and ultimately decay. Can anyone seriously claim that the finiteness of the perturbative S-matrix elements of a badly divergent perturbation series settles the question of whether the soft graviton state produced in this process is a normalizable state in Fock space? The only non-perturbative model of gravitational scattering in Minkowski space, of which we are aware is the large N limit of Matrix Theory[20]. In this model, soft gravitons correspond to very small matrix blocks, which carry very small transverse momentum. As N goes to infinity, it’s clear that we must examine the question of whether the unitary scattering matrix of the finite N theory decouples from the states with infinite numbers of such small blocks. There is absolutely no indication that it will do so.
Update (4/2019): From the introduction to a recent textbook on string theory:
A big “hole” in string theory has been its perturbative (only) definition. With the advent of nonperturbative dualities, it was hoped that this shortcoming can be bypassed.Although the nonperturbative dualities have shed light in many obscure corners of string theory (obscured by strong-coupling physics), they never managed to bypass the Planck barrier. The Planck scale is always duality invariant, and any dual description is well defined for energies well below that Planck scale. We have no clue from string theory what happens near or above the Planck scale, as the relevant physics looks nonperturbative from any point of view.
