Dear Adrien, The thermal field represents the fluctuactions, which will give the ensamble averages, that is Boltzmann distribution for a single cell. It depends in temperature, damping and timestep. Therefore, from your email is not very easy to conclude which are the values for that. But it is easy to verify that mumax follow the Boltzmann distribution.
To check that one can run a cell with only uniaxial anisotropy (no exchange no dipolar field). A script like this would do that:
SetMesh(2, 2, 2, 4e-09, 4e-09, 4e-09, 0, 0, 0)
Msat = 1e6
Aex = 0
alpha = 0.01
anisU = vector(0, 0, 1)
Ku1=1e5
m = RandomMag()
FixDt = 1e-14
SetSolver(4)
EnableDemag=False
DefRegionCell(1,1,1,1)
Temp=400
tableAdd(m.Region(1))
tableAdd(B_therm)
tableautosave(1e-12)
Run(100e-09)
With this data the average value of mz for 400 K should be according to Boltzmann distribution (the ratio of Ku*V/kbT is 1.159 in that case) 0.601643. For a single run I get a value of 0.60519, which is in good agreement. For 100 K, the theoretical value should be 0.844196 and the calculated value is 0.84482. Therefore, the fields are correctly calculated and have to be interpreted in that sense.
Regarding your cell, it is small (I guess 5x3x1nm3) and therefore it will have large fluctuations. A particle like that would be superparamagnetic if it would be an isolated nanoparticle. In the micromagnetic case exchange coupling with neighboring cells will allow to have a net magnetization in the sample but for example temperature is at the end enable to overcome the exchange beyond Curie temperature. I don't think there is anything wrong in your calculations unless you used an unrealistic value for the damping or the timestep (timesteps seem correct). The events of depinning will depended also on the depth of your pinning potential.