Your idea can be made to work but it involves some calculations. Let's write vectors in $\mathbb{R}^3$ as column vectors and choose $X \colon U \rightarrow \mathbb{R}^3$ such that $U \subseteq \mathbb{R}^2$ is an open set and $X$ is a parametrization of almost all of $\mathbb{S}^2$. For example, $X$ can be defined using spherical coordinates but the specific form of $X$ does not matter. Then $Y \colon U \rightarrow \mathbb{R}^3$ given by
$$ Y(p) = \underbrace{f(X(p))}_{:=h(p)} X(p) $$
is a parametrization of almost all of $M$. First of all, note that $\| X(p) \|^2 = X^T \cdot X = 1$ for all $p \in U$. Differentiating this identity, we obtain
$$ X^T \cdot dX = 0 $$
(where $dX$ is the $3 \times 2$ matrix representing the differential).
Now, using the chain rule, we have
$$ dY = (X \cdot dh + h dX) $$
and so
$$ dY^T \cdot dY = \left( \nabla h \cdot X^T + h dX^T\right) \cdot \left(X \cdot \left( \nabla h \right)^T + h dX\right) = \nabla h \cdot \left( \nabla h \right)^T + h \left( \nabla h \cdot \underbrace{X^T \cdot dX}_{0} + \underbrace{dX^T \cdot X}_{0} \cdot \left( \nabla h \right)^T \right) + h^2 (dX^T \cdot dX) =
(\nabla h) \cdot \left( \nabla h \right)^T + h^2 (dX^T \cdot dX). $$
Using the matrix determinant lemma we get
$$ \det \left( Y^T \cdot Y \right) = \left( 1 + \left( \nabla h \right)^T \left( h^2 \left( dX^T \cdot dX \right)\right)^{-1} \cdot \nabla h \right) h^4 \det \left( X^T \cdot X \right) = \left( h^4 + h^2 \left< \nabla h, \left( dX^T \cdot dX \right)^{-1} \nabla h \right>\right) \det \left( X^T \cdot X \right). $$
Note that $\left( dX^T \cdot dX \right)^{-1}$ is a positive definite symmetric matrix and so $\left< \nabla h, \left( dX^T \cdot dX \right)^{-1} \nabla h \right> \geq 0$. Together with the fact that $h \geq 1$, we can conclude that
$$ \det \left( Y^T \cdot Y \right) \geq \det \left(X^T \cdot Y \right). $$
Finally,
$$ \textrm{Area}(M) = \iint_{U} \sqrt{\det \left( Y^T \cdot Y \right)} \geq \iint_{U} \sqrt{\det \left( X^T \cdot X \right)} = \textrm{Area}(\mathbb{S}^2). $$
