Willmore energy in Riemannian Schwarzschild manifold The $3$-dimensional Riemannian Schwarzschild manifold $(M,\bar{g})$ is given by $M=[s_0,\infty)\times\mathbb{S}^2$ with 
\begin{align}
\bar{g}=\frac{1}{1-\frac{2m}{s}}ds\otimes ds+s^2g_{\mathbb{S}^2}
\end{align}
where $m>0$ and $g_{\mathbb{S}^2}$ is the standard round metric on the unit $2$-sphere. By a change of variable, it can be written as a warped product $M=[0,\infty)\times\mathbb{S}^2$ with
\begin{align}
\bar{g}=dr\otimes dr+u(r)^2g_{\mathbb{S}^2}
\end{align}
where $u:[0,\infty)\to\mathbb{R}$ is the smooth warping function satisfying
\begin{align}
u'(r):=\frac{du}{dr}=\sqrt{1-\frac{2m}{u(r)}}
\end{align}
In this question I am interested in the Willmore energy
\begin{align}
\int_{\Sigma_r}H^2d\mu
\end{align}
(my mean curvature here is $H=\lambda_1+\lambda_2$ w/o dividing by 4) of the slices $\{r\}\times\mathbb{S}^2$ in $(M,\bar{g})$. 
Since $(M,\bar{g})$ is a warped product, I can compute that
\begin{align}
\int_{\Sigma_r}H^2d\mu&=16\pi u'(r)^2 \\
&=16\pi\left(1-\frac{2m}{u(r)}\right)
\end{align}
which is dependent on $r$. 
On the other hand, it is also known that $(M,\bar{g})$ can be written as $(\mathbb{R}^3\setminus B_{2m}(0),\bar{g})$ with
\begin{align}
\bar{g}=\left(1+\frac{m}{2|x|}\right)^4\delta
\end{align}
where $\delta$ is the standard Euclidean flat metric, and $|\cdot|$ is the Euclidean norm. This shows that $\bar{g}$ is conformal to $\delta$. Now, since Willmore energy is a conformal invariant, and since $\Sigma_r$ is a round sphere, it follows from the standard result by Willmore himself that
\begin{align}
\int_{\Sigma_r}H^2d\mu=16\pi
\end{align}
which clearly is an absolute constant. 
In summary, I obtain two different answers from different reasonings. At least one of them must contain some fallacy which I fail to notice. Hence, I would like to ask for an explanation on this. 
Any comment and answer are welcomed and greatly appreciated. 
 A: Your mistake is in the interpretation of conformal invariance.  The Willmore energy of a surface in $\mathbb{R}^3$ is conformally invariant in the sense that it is invariant under Möbius transformations.
A better way to describe the conformal invariance of the Willmore energy is as follows.  Let $\Sigma$ be a hypersurface in a Riemannian manifold $(M^3,g)$.  Define the Willmore energy of $\Sigma$ by
$$ W(\Sigma,g) := \int_\Sigma \lvert A_0\rvert^2\,d\mu, $$
where $A_0$ is the trace-free part of the second fundamental form.  This is conformally invariant, in the sense that $W(\Sigma,u^2g)=W(\Sigma,g)$ for any positive function $u$ on $M$, for the simple reason that $\lvert A_0\rvert_{u^2g}^2\,d\mu_{u^2g}=\lvert A_0\rvert_g^2\,d\mu_g$ (this uses the assumption $\dim M=3$).
Now, by the Gauss equation,
$$ \label{e} \tag{$\ast$} W(\Sigma,g) + 4\pi\chi(\Sigma) = \frac{1}{2}\int_\Sigma H^2\,d\mu + \int_\Sigma \left( R - 2\mathrm{Ric}_{nn} \right)\,d\mu , $$
where $R$ is the scalar curvature of $g$ and $\mathrm{Ric}_{nn}$ is the normal component of the Ricci tensor of $g$.  In your situation, $R=0$ but $\mathrm{Ric}_{nn}\not=0$, which explains the discrepancy.
(Note that \eqref{e} also gives the familiar formula for the Willmore energy of a surface $\Sigma\subset\mathbb{R}^3$ after pulling it back to $S^3$ by stereographic projection.)
