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.