Play with a Gaussian Random Set in the Frequency Domain to Obtain Desired Effect in the Time Domain
You're asking to do a localized operation in time using the Frequency Domain.  

It's going to be not elegant, really.

Here what you can do:


Define the input signal in frequency domain as $ X \left[ k \right] $.
Multiply by Linear Phase in Frequency Domain to apply circular shift in time domain.
This means samples 1-90 will become 11-100 and 91-100 will become 1-10. Lets' call this multiplied signal $ \hat{X} \left[ k \right] $.
Pre calculate the DFT of a signal which is 1 for the first 10 samples and then zero for samples 11-100. Lets call it $ H \left[ k \right] $. This is basically a windows on the first 10 samples. In order to extract the 10 first samples in time domain we would multiply by it. In the frequency domain it means we will apply Convolution. The result of the convolution will be a frequency domain signal which is equivalent of the DFT of only the 10 first samples padded with 90 zeros. Keep this result on side. Let's Call it $ Y \left[ k \right] = \hat{X} \left[ k \right] \ast H \left[ k \right] $.
Generate noise on frequency domain which is with the properties needed of generate what's needed on time domain. Convolve the generated noise with $ H \left[ k \right] $ which will again extract a signal which is equivalent to only 10 first samples in time domain. From this signal, lets call it $ Z \left[ k \right] $ subtract the signal $ Y \left[ k \right] $.
Add the result to $ \hat{X} \left[ k \right] $. Now the signal $ \hat{X} \left[ k \right] + Z \left[ k \right] - Y \left[ k \right] $ is what you're after.


Pay attention that doing this in Fourier domain might be much less efficient than doing it in time domain.
As you need to apply convolution twice instead of doing 2 DFT's.