Are perturbative and non-perturbative QCD both signs of new physics?

I was studying about quarkonia systems and reached this page at CERN Courier. Here, I came across the following text:

While the failure to reproduce an experimental observable that is perturbatively calculable in the electroweak or strong sector would be interpreted as a sign of new physics, phenomena requiring a non-perturbative treatment – such as those related to the long-distance regime of the strong force – are less likely to trigger an immediate reaction.

What does this exactly mean? Can somebody roughly sum it up, using some example, if possible?

An interesting example relevant to the discussion is the study of CP violation in B meson decays. One is interested in extracting the parameters of the Standard Model responsible for mixing quark flavours (they are gathered in a matrix called the CKM matrix). In practice, one seek to measure a phase angle originating in the coefficients of this matrix. But QCD can contribute an extra phase, and some of that is non-perturbative. As a result, the "golden" channels are those where QCD does not contribute any phase. For example, $b\to c\bar{c}s$. So if one would see a strong deviation from the Standard Model in this golden channel, people would be very confident that there is new physics. But on the contrary, a deviation in a channel with a sizeable QCD contribution, such as $b\to c\bar{c}d$, would raise less of an eyebrow.
• There are various computational hurdles that can't be easily dwarfed as I understand it. The biggest problem is the evaluation of the so-called fermion determinant. Most of the flops comes from inverting a matrix using iterative methods and the computing cost is then proportional to the condition number, which is horrible because the smallest eigenvalue is the mass $m$ of the quark, i.e. it scales as $1/m$. – user154997 Jul 1 '17 at 19:09
• @EmilioPisanty There is that, indirectly. If $N$ is the number of lattice steps in each dimension, then a naive computation of the fermion determinant I talked about scales as $N^{12}$. A cost to be paid at each iteration of a Monte-Carlo integration. But a clever workaround avoid this cost, leading to the point I made above. – user154997 Jul 1 '17 at 22:17