In general, $U_s$ is not recoverable from $H$ and $U_A$.
Consider the following example in which 2 different $R$'s, $R1$ and $R2$, have different $U_s$'s while having the same $U_A$.
MATLAB output:
>> disp(H)
1 0 0
0 2 0
>> disp(R1)
2 1 1
1 2 1
1 1 2
>> [U_s_R1,lambda_R1]=eig(R1)
U_s_R1 =
0.408248290463863 0.707106781186547 0.577350269189626
0.408248290463863 -0.707106781186547 0.577350269189625
-0.816496580927726 0 0.577350269189626
lambda_R1 =
0.999999999999999 0 0
0 1.000000000000000 0
0 0 3.999999999999999
>> disp(R2)
9.999999999999929 4.999999999999965 0.984522053823275
4.999999999999965 9.999999999999929 0.984522053823275
0.984522053823275 0.984522053823275 9.999999999999929
>> [U_s_R2,lambda_R2]=eig(R2)
U_s_R2 =
0.707106781186548 0.177730756491407 0.684406150028616
-0.707106781186547 0.177730756491407 0.684406150028616
0 -0.967896459542024 0.251349246280978
lambda_R2 =
4.999999999999966 0 0
0 9.638432711095330 0
0 0 15.361567288904492
>> [U_A_R1,lambda_A_R1]=eig(H*R1*H')
U_A_R1 =
-0.957092026489053 0.289784148688430
0.289784148688430 0.957092026489053
lambda_A_R1 =
1.394448724536011 0
0 8.605551275463990
>> [U_A_R2,lambda_A_R2]=eig(H*R2*H')
U_A_R2 =
-0.957092026489053 0.289784148688430
0.289784148688430 0.957092026489053
lambda_A_R2 =
6.972243622680004 0
0 43.027756377319641
As can be seen, U_A_R1 = U_A_R2, but U_s_R1 shares only one column with U_s_R2, i.e., R1 has only one eigenvector in common with R2.