Sampling of band-limited white noise
There are two things to consider here: the noise power after sampling, and the fact whether the sampled noise is uncorrelated or not. Note that the autocorrelation function of the sampled noise equals the sampled autocorrelation function of the continuous-time noise. For a wide-sense stationary (WSS) band-limited white noise signal $X(t)$ with bandwidth $B$ and one-sided noise density $N_0$, the autocorrelation function is given by

$$R_X(\tau)=BN_0\,\text{sinc}(2B\tau)\tag{1}$$

where I use $\text{sinc}(x)=\sin(\pi x)/(\pi x)$.

If you define a discrete time process $Y[n]=X(nT)$, where $f_s=1/T$ is the sampling frequency, then its autocorrelation function is given by

$$R_Y[k]=R_X(kT)=BN_0\,\text{sinc}(2BkT)\tag{2}$$

Since the power of the discrete-time noise $Y[n]$ is given by $R_Y[0]=BN_0$, it is obvious that the sampling frequency has no influence on the noise power. So no matter how fast you sample the original continuous-time noise, the power of the resulting discrete-time noise always equals the power of the noise before sampling.

Now let's have a look at the correlation between the noise samples. The autocorrelation function $R_X(\tau)$ has zeros at

$$\tau_i=\frac{i}{2B},\quad i=1,2,\ldots$$

which for $B=10\,\text{kHz}$ gives $\tau_i=i\cdot 50\mu s$ (this is the small numerical error in Dilip Sarwate's post you referred to in your question). This means that if you choose a sampling interval which is an integer multiple of $50\mu s$, the resulting sampled noise will be uncorrelated. This corresponds to sampling frequencies $f_s=2B/i$, $i=1,2,\ldots$, i.e. $20\,\text{kHz}$, $10\,\text{kHz}$, $6.67\,\text{kHz}$, etc.

So with a sampling frequency of $20\,\text{kHz}$ you do get uncorrelated noise samples.