I wasn't able to find a source to cite for the correct formulae, but it turns out that it's not that hard to work out the answer directly using differential geometry.
First, a little notation: Let $u:S^2\to\mathbb{R}^3$ be the usual inclusion mapping. A support function $p:S^2\to\mathbb{R}$ defines a family of planes $u\cdot x - p(u) = 0$, and the formula for the envelope of this family of planes is
$$
x = p\,u + \nabla p,\tag1
$$
where $(\nabla p)(u)\in T_uS^2\subset\mathbb{R}^3$ is the gradient of $p$ computed with respect to the standard (round) metric on $S^2$ induced by the inclusion $u:S^2\to\mathbb{R}^2$. Then the area $2$-form induced on $S^2$ by the inclusion $x:S^2\to\mathbb{R}^3$ is simply given by the scalar product
$$
\mathrm{d}A = \tfrac12\,u\cdot(\mathrm{d}x\times \mathrm{d}x).
$$
Substituting (1) into this formula, we find that
$$
\mathrm{d}A = \bigl(p^2 - p\Delta p + \det(\nabla^2p)\bigr)\,\mathrm{d}\sigma,
$$
where $\mathrm{d}\sigma$ is the standard area form on $S^2$, $\nabla^2p$ is the second covariant derivative of $p$ (often called the Hessian of $p$), and $\Delta p$ is the negative trace of $\nabla^2p$, i.e., the geometer's Laplacian (which is a positive operator).
Meanwhile, for any immersion, $x:S^2\to\mathbb{R}^3$, Stokes' Theorem implies that the (algebraic) volume enclosed by $\Sigma = x(S^2)$ is given by
$$
V = \frac16\int_{S^2} x\cdot (\mathrm{d}x\times \mathrm{d}x),
$$
and computation using (1) above yields
$$
x\cdot (\mathrm{d}x\times \mathrm{d}x)
= 2p\bigl(p^2 - p\Delta p + \det(\nabla^2p)\bigr)\,\mathrm{d}\sigma
= 2p\,\mathrm{d}A.$$
Hence
$$
V = \frac13 \int_\Sigma p\,\,\mathrm{d}A,
$$
in accordance with the source the OP cites.
This suggests that the area formula that the OP doubted might be corrected to $A = \int_{S^2} p^2\,\mathrm{d}\sigma$. (At least this would have the correct homogeneity with respect to scaling.) However, that is not the case. The correct formula is found to be as follows: Using harmonic analysis to expand $p$ in eigenfunctions of $\Delta$,
i.e., $p = p_0 + p_1 + \cdots + p_k + \cdots$
where $\Delta p_k = k(k{+}1)\,p_k$, one finds that
$$
\int_{S^2} p^2\,\mathrm{d}\sigma = \sum_{k=0}^\infty \int_{S^2} {p_k}^2\,\mathrm{d}\sigma,
$$
while
$$
A = \int_{S^2} \bigl(p^2 - p\Delta p + \det(\nabla^2p)\bigr)\,\mathrm{d}\sigma
= \sum_{k=0}^\infty \frac{(1{-}k)(2{+}k)}2\int_{S^2} {p_k}^2\,\mathrm{d}\sigma.
$$
Thus, we only get $A\le \int_{S^2} p^2\,\mathrm{d}\sigma$, with equality if and only if $p$ is constant (i.e., $p = p_0$).
Here is how one can see that the above formula for $A:C^\infty(S^2)\to\mathbb{R}$ is correct: Note that, as a function of $p\in C^\infty(S^2)$, the integral $A$ is a quadratic form that is invariant under the action of $\mathrm{SO}(3)$. Now, as a representation of $\mathrm{SO}(3)$, $C^\infty(S^2)$ is a sum of the irreducible finite dimensional representations $\mathcal{H}_k$ $(k\ge0)$, where $\mathcal{H}_k$ is the space of eigenfunctions satisfying $\Delta u = k(k{+}1)u$ and is also the space of functions got by restricting the harmonic polynomials of degree $k$ on $\mathbb{R}^3$ to the $2$-sphere. Each representation $\mathcal{H}_k$ occurs with multiplicity $1$. Hence, the invariance of $A$ under the action of $\mathrm{SO}(3)$ implies that $\mathcal{H}_k$ and $\mathcal{H}_j$ are $A$-orthogonal when $j\not=k$. This implies that
$$
A = \int_{S^2} \bigl(p^2 - p\Delta p + \det(\nabla^2p)\bigr)\,\mathrm{d}\sigma
= \sum_{k=0}^\infty \mu_k \int_{S^2} {p_k}^2\,\mathrm{d}\sigma.
$$
for some constants $\mu_k$. To evaluate the constant $\mu_k$, it suffices to compute $A$ for $p = p_k = \mathrm{Re}\bigl((x+iy)^k\bigr)$ and compare the result with $\int_{S^2}{p_k}^2\,\mathrm{d}\sigma$. Using standard spherical coordinates, this is a straightforward exercise in vector calculus, yielding the values of $\mu_k$ that are displayed above.
