No, Bayesian decision theory does not allow you to reason in that way.

Suppose the amounts in the envelopes are A and B. A Bayesian would say:

Prob(I have A) = 1/2 and Prob(I have B) = 1/2

The expected gain from switching is:

Prob(I have A)x(what I'd gain from switching = B-A) + Prob(I have B)x(what I'd gain from switching = A-B)

which is: 1/2 x (B-A) + 1/2 x (A-B) = 0