$\newcommand{\R}{\mathbb R}\newcommand{\SSS}{\mathbb S}\newcommand{\tS}{\tilde S}\newcommand{\tR}{\tilde R}$This is to detail and formalize the nice counterexample outlined in fedja's comment.
First, let us recapitulate the counterexample informally. Let $d=3$. Consider the point $P:=(0,0,1)$ on the unit sphere $\SSS_2$. In the plane $\Pi:=\{(x,y,z)\in\R^3\colon z=1\}$, tangent to $\SSS_2$ at $P$, take a small thin diamond shape $\tS$ centered at $P$. Take then a very thin rectangle $\tR$ inscribed into the diamond $\tS$ along its bigger diagonal. Let $R$ and $S$ be the images of $\tR$ and $\tS$, respectively, under the
central projection (with the center at the origin) onto the sphere $\SSS_2$.
By symmetry, the centroids $EX_R$ and $EX_S$ of $R$ and $S$ will then be on the $z$-axis, so that $EX_R=(0,0,z_R)$ and $EX_S=(0,0,z_S)$ for some $z_R$ and $z_S$ in the interval $(0,1)$, and hence $|EX_R|=z_R$ and $|EX_S|=z_S$. Note that $z_R$ and $z_S$ are $\approx1$. So, it makes sense to compare $1-z_R$ and $1-z_S$.
For points $(x,y,z)\in\SSS_2$ close to $P$, we have $1-z=1-\sqrt{1-x^2-y^2}\sim(x^2+y^2)/2$. So, comparing the averages of $z$ over $R$ and $S$ is asymptotically equivalent to comparing the averages of $x^2+y^2$ over $R$ and $S$, and hence to comparing the averages of $x^2+y^2$ over the flat regions $\tR$ and $\tS$, since the central projection from the sphere $\SSS_2$ onto the plane $\Pi$ distorts the lengths negligibly near $P$.
It remains to compare the averages of $x^2+y^2$ over the flat regions $\tR$ and $\tS$, which is straightforward.
Let us now formalize the reasoning. For $h\downarrow0$, consider the diamond shape
\begin{equation*}
\tS:=\Big\{(x,y,1)\in\R^3\colon \frac{|x|}h+\frac{|y|}{h^2}\le1\Big\}
\end{equation*}
in the plane $\Pi$ and its narrow part
\begin{equation*}
\tR:=\Big\{(x,y,1)\in\R^3\colon \frac{|x|}h+\frac{|y|}{h^2}\le1,\, |y|\le h^3\Big\}.
\end{equation*}
Let $R$ and $S$ be the images of $\tR$ and $\tS$, respectively, under the
central projection (with the center at the origin) onto the sphere $\SSS_2$. So,
\begin{equation*}
S=\big\{r(x,y)\colon(x,y)\in\rho_S\big\}\quad\text{and}\quad R=\big\{r(x,y)\colon(x,y)\in\rho_R\big\},
\end{equation*}
where
\begin{equation*}
r(x,y):=\frac{(x,y,1)}{\sqrt{x^2+y^2+1}},
\end{equation*}
\begin{equation*}
\rho_S:=\Big\{(x,y)\in\R^2\colon\frac{|x|}h+\frac{|y|}{h^2}\le1\Big\},
\end{equation*}
\begin{equation*}
\rho_R:=\Big\{(x,y)\in\R^2\colon\frac{|x|}h+\frac{|y|}{h^2}\le1,\,|y|\le h^3\Big\}.
\end{equation*}
So, the regions $S$ and $R$ on the sphere are parametrized by parameters $x,y$, with $(x,y)\in\rho_S$ and $(x,y)\in\rho_R$, respectively.
With this parametrization, the spherical area element is
\begin{equation*}
dA(x,y)=\|r_x(x,y)\times r_y(x,y)\|\,dx\,dy \\
=\frac{dx\,dy}{(x^2+y^2+1)^{3/2}}\sim dx\,dy
\end{equation*}
for $(x,y)\in\rho_S$, where $r_x$ and $r_y$ are the partial derivatives of the vector function $r$ with respect to $x$ and $y$; $\times$ is the cross product; and $\|\cdot\|$ is the length. This result for $dA(x,y)$ is also easy to obtain geometrically.
Introducing the $z$-coordinate of $r(x,y)$,
\begin{equation*}
Z(x,y):=\frac1{\sqrt{x^2+y^2+1}},
\end{equation*}
we see that
\begin{equation*}
1-Z(x,y)\sim\frac{x^2+y^2}2
\end{equation*}
for $(x,y)\in\rho_S$ and hence for $(x,y)\in\rho_R$.
So,
\begin{equation*}
\iint_{\rho_S}dA(x,y)
\sim\iint_{\rho_S}dx\,dy=2h^3,
\end{equation*}
\begin{equation*}
\iint_{\rho_S}(1-Z(x,y))dA(x,y)
\sim\frac12\,\iint_{\rho_S}(x^2+y^2)\,dx\,dy \\
=\frac16\,h^5(1+h^2)\sim\frac{h^5}6,
\end{equation*}
whence
\begin{equation*}
1-z_S=\dfrac{\iint_{\rho_S}(1-Z(x,y))dA(x,y)}{\iint_{\rho_S}dA(x,y)}
\sim\frac{h^5}{6}/(2h^3)=\frac{h^2}{12}. \tag{2}\label{2}
\end{equation*}
Also,
\begin{equation*}
\iint_{\rho_R}dA(x,y)\sim4\int_0^{h^3}dy\,\int_0^{h-O(h^2)} dx\sim4h^4,
\end{equation*}
\begin{equation*}
\iint_{\rho_R}(1-Z(x,y))dA(x,y)
\sim\frac12\,\iint_{\rho_R}(x^2+y^2)\,dx\,dy \\
=\frac42\,\int_0^{h^3}dy\,\int_0^{h-O(h^2)}(x^2+y^2)\, dx\sim\frac{2h^6}3,
\end{equation*}
whence
\begin{equation*}
1-z_R=\dfrac{\iint_{\rho_R}(1-Z(x,y))dA(x,y)}{\iint_{\rho_R}dA(x,y)}
\sim\frac{2h^6}3/(4h^4)=\frac{h^2}6. \tag{3}\label{3}
\end{equation*}
By \eqref{2} and \eqref{3}, for all small enough $h>0$ we have $1-z_R>1-z_S$, that is $z_S>z_R$, that is $|E(X_S)|>|E(X_R)|$, which disproves the conjecture $|E(X_S)|\le|E(X_R)|$.
