Part 1
The differential forms approach is indeed very powerful. What Hestenes points out in his From Clifford Algebra to Geometric Calculus is that to give a complete treatment of differential geometry of manifolds you need various structures. In the book, you will find an alternative. The starting point (as was pointed out above) is the notion of a vector manifold.
A vector manifold is an abstract set contained within an infinite dimensional abstract algebra. This can be given different interpretations.
There are special elements of this algebra called vectors and others called pseudoscalars. The set of elements that define a vector manifold is a set of vectors. These vectors generate tangent space that are not part of the set. This is customary: one often invokes tangent spaces on manifolds. Notice that the vectors which define the vector manifold and the vectors which form the tangent space are very different. For visualization purposes, the elements of a vector manifold are called "points." The space tangent to a point generates a tangent algebra; this algebra contains ONE element called its pseudoscalar.
The function $I_m(x)$, which is pseudoscalar-valued and takes as arguments "points" of the vector manifold, characterizes the vector manifold. If this function is differentiable then one says that the vector manifold is differentiable. If so all the differential geometry of a vector manifold can be carried out using only $I_m(x)$. Here, $m$ is the dimension of the tangent spaces, this also defines the dimension of the vector manifold.
So the idea behind the definition is the following. If you have a manifold, its points lack algebraic structure, and so one imposes other structures as needed. If, on the other hand, you start with a vector manifold which is algebraically rich, then no further structures are needed later on. A manifold can be defined as a space which is isomorphic to a vector manifold. This isomorphism can be thought of as a mere stripping of the algebraic structure. A manifold can be treated as a completely abstract object, but for the purposes for which it was constructed, a geometrical interpretation is rewarding. This is also true for vector manifolds.
Some may think that vector manifolds are embedded in a Euclidean space or that one needs a metric, but this is not true. The main reason could be (this is my opinion) the nomenclature of GA. For example a vector manifold sounds like a special kind of manifold. Another case is the inner product. This product is defined algebraically and does not need a metric. It can be interpreted metrically and thus lead to very important developments, but it is only an interpretation. I use "only" in the sense that a metric does not define the abstract algebra or anything about it.
Part 2
Differential forms can live in the GA framework as follows. There is a unique $k$-vector for every $k$-form. A one-to-one correspondence.
Take the following example. Suppose you want to manipulate an expression before integrating. You do so and right before you integrate, you "multiply" by a differential. Then you integrate.
You can "multiply" by the differential from the very beginning, but you do not need to do so and would probably not do so. The difference between a $k$-vector and a $k$-form is somewhat similar. If you already know what a $k$-form is, you could think of a $k$-vector as a $k$-form without the differential. If you don't know what a $k$-form is, you can study $k$-vectors and some of their mathematics and right before you need to integrate a $k$-vector, you turn it into a $k$-form. (I point this out because it is more or less how Hestenes proceeds in the book I mentioned above: forms are only strictly needed when integration is to be carried out.)
What I am pictorially trying to answer with the above example is "are these frameworks strictly equivalent?" When scalar-valued integrals are considered, the answer is YES. GA can also handle vector-valued integrals or in general multivector integrals. And integrals are Riemannian! Outside of integration theory, GA is more general. For example, the exterior derivative of differential forms has a counterpart in GA. Both are grade raising. GA also has a grade lowering derivative, but in differential forms, this can only be achieved with a metric. When comparing the two, one might get confused and think GA also needs a metric for this. This is no true. Furthermore, the sum of these grade raising and lowering operators is in actuality the fundamental derivative of GA this derivative is invertible while the exterior derivative is in general not.
The overall answer to your question is geometric algebra (technically geometric calculus) is more general.
Comment.
There are also other alternatives to geometric calculus which do not use the concept of a vector manifold. They are also more general by treating any manifold and use Clifford algebra structures and do not use coordinates. Some other approaches exist which use coordinates here and there but also use this algebra structure and are more general.
I just want to point out that GA can be used to make covariant multivectors (or differential forms) on $\mathbb R^n$ without forcing a metric onto it. In other words, the distinction between vectors and covectors (or between $\mathbb R^n$ and its dual) can be maintained.
This is done with a pseudo-Euclidean space $\mathbb R^{n,n}$.
Take an orthonormal set of spacelike vectors $\{\sigma_i\}$ (which square to ${^+}1$) and timelike vectors $\{\tau_i\}$ (which square to ${^-}1$). Define null vectors
$$\Big\{\nu_i=\frac{\sigma_i+\tau_i}{\sqrt2}\Big\}$$
$$\Big\{\mu_i=\frac{\sigma_i-\tau_i}{\sqrt2}\Big\};$$
they're null because
$${\nu_i}^2=\frac{{\sigma_i}^2+2\sigma_i\cdot\tau_i+{\tau_i}^2}{2}=\frac{(1)+2(0)+({^-}1)}{2}=0$$
$${\mu_i}^2=\frac{{\sigma_i}^2-2\sigma_i\cdot\tau_i+{\tau_i}^2}{2}=\frac{(1)-2(0)+({^-}1)}{2}=0.$$
More generally,
$$\nu_i\cdot\nu_j=\frac{\sigma_i\cdot\sigma_j+\sigma_i\cdot\tau_j+\tau_i\cdot\sigma_j+\tau_i\cdot\tau_j}{2}=\frac{(\delta_{i,j})+0+0+({^-}\delta_{i,j})}{2}=0$$
and
$$\mu_i\cdot\mu_j=0.$$
So the spaces spanned by $\{\nu_i\}$ or $\{\mu_i\}$ each have degenerate quadratic forms. But the dot product between them is non-degenerate:
$$\nu_i\cdot\mu_i=\frac{\sigma_i\cdot\sigma_i-\sigma_i\cdot\tau_i+\tau_i\cdot\sigma_i-\tau_i\cdot\tau_i}{2}=\frac{(1)-0+0-({^-}1)}{2}=1$$
$$\nu_i\cdot\mu_j=\frac{\sigma_i\cdot\sigma_j-\sigma_i\cdot\tau_j+\tau_i\cdot\sigma_j-\tau_i\cdot\tau_j}{2}=\frac{(\delta_{i,j})-0+0-({^-}\delta_{i,j})}{2}=\delta_{i,j}$$
Of course, we could have just started with the definition that $\mu_i\cdot\nu_j=\delta_{i,j}=\nu_i\cdot\mu_j$, and $\nu_i\cdot\nu_j=0=\mu_i\cdot\mu_j$, instead of going through "spacetime".
The space $V$ will be generated by $\{\nu_i\}$, and its dual $V^*$ by $\{\mu_i=\nu^i\}$. You can take the dot product of something in $V^*$ with something in $V$, which will be a differential 1-form. You can make contravariant multivectors from wedge products of things in $V$, and covariant multivectors from wedge products of things in $V^*$.
You can also take the wedge product of something in $V^*$ with something in $V$.
$$\mu_i\wedge\nu_i=\frac{\sigma_i\wedge\sigma_i+\sigma_i\wedge\tau_i-\tau_i\wedge\sigma_i-\tau_i\wedge\tau_i}{2}=\frac{0+\sigma_i\tau_i-\tau_i\sigma_i-0}{2}=\sigma_i\wedge\tau_i$$
$$\mu_i\wedge\nu_j=\frac{\sigma_i\sigma_j+\sigma_i\tau_j-\tau_i\sigma_j-\tau_i\tau_j}{2},\quad i\neq j$$
What does this mean? ...I suppose it could be a matrix (a mixed variance tensor)!
A matrix can be defined as a bivector:
$$M = \sum_{i,j} M^i\!_j\;\nu_i\wedge\mu_j = \sum_{i,j} M^i\!_j\;\nu_i\wedge\nu^j$$
where each $M^i_j$ is a scalar. Note that $(\nu_i\wedge\mu_j)\neq{^-}(\nu_j\wedge\mu_i)$, so $M$ is not necessarily antisymmetric. The corresponding linear function $f:V\to V$ is (with $\cdot$ the "fat dot product")
$$f(x) = M\cdot x = \frac{Mx-xM}{2}$$
$$= \sum_{i,j} M^i_j(\nu_i\wedge\mu_j)\cdot\sum_k x^k\nu_k$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i\mu_j-\mu_j\nu_i}{2}\cdot\nu_k$$
$$= \sum_{i,j,k} M^i_jx^k\frac{(\nu_i\mu_j)\nu_k-\nu_k(\nu_i\mu_j)-(\mu_j\nu_i)\nu_k+\nu_k(\mu_j\nu_i)}{4}$$
(the $\nu$'s anticommute because their dot product is zero:)
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i\mu_j\nu_k+\nu_i\nu_k\mu_j+\mu_j\nu_k\nu_i+\nu_k\mu_j\nu_i}{4}$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\mu_j\nu_k+\nu_k\mu_j)+(\mu_j\nu_k+\nu_k\mu_j)\nu_i}{4}$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\mu_j\cdot\nu_k)+(\mu_j\cdot\nu_k)\nu_i}{2}$$
$$= \sum_{i,j,k} M^i_jx^k\frac{\nu_i(\delta_{j,k})+(\delta_{j,k})\nu_i}{2}$$
$$= \sum_{i,j,k} M^i_jx^k\big(\delta_{j,k}\nu_i\big)$$
$$= \sum_{i,j} M^i_jx^j\nu_i$$
This agrees with the conventional definition of matrix multiplication.
In fact, it even works for non-square matrices; the above calculations work the same if the $\nu_i$'s on the left in $M$ are basis vectors for a different space. A bonus is that it also works for a non-degenerate quadratic form; the calculations don't rely on ${\mu_i}^2=0$, nor ${\nu_i}^2=0$, but only on $\nu_i$ being orthogonal to $\nu_k$, and $\mu_j$ being reciprocal to $\nu_k$. So you could instead have $\mu_j$ (the right factors in $M$) be in the same space as $\nu_k$ (the generators of $x$), and $\nu_i$ (the left factors in $M$) in a different space. A downside is that it won't map a non-degenerate space to itself.
I admit that this is worse than the standard matrix algebra; the dot product is not invertible, nor associative. Still, it's good to have this connection between the different algebras. And it's interesting to think of a matrix as a bivector that "rotates" a vector through the dual space and back to a different point in the original space (or a new space).
Speaking of matrix transformations, I should discuss the underlying principle for "contra/co variance": that the basis vectors may vary.
We want to be able to take any (invertible) linear transformation of the null space $V$, and expect that the opposite transformation applies to $V^*$. Arbitrary linear transformations of the external $\mathbb R^{n,n}$ will not preserve $V$; the transformed $\nu_i$ may not be null. It suffices to consider transformations that preserve the dot product on $\mathbb R^{n,n}$. One obvious type is the hyperbolic rotation
$$\sigma_1\mapsto\sigma_1\cosh\phi+\tau_1\sinh\phi={\sigma_1}'$$
$$\tau_1\mapsto\sigma_1\sinh\phi+\tau_1\cosh\phi={\tau_1}'$$
$$\sigma_2={\sigma_2}',\quad\sigma_3={\sigma_3}',\quad\cdots$$
$$\tau_2={\tau_2}',\quad\tau_3={\tau_3}',\quad\cdots$$
(or, more compactly, $x\mapsto\exp(-\sigma_1\tau_1\phi/2)x\exp(\sigma_1\tau_1\phi/2)$ ).
The induced transformation of the null vectors is
$${\nu_1}'=\frac{{\sigma_1}'+{\tau_1}'}{\sqrt2}=\exp(\phi)\nu_1$$
$${\mu_1}'=\frac{{\sigma_1}'-{\tau_1}'}{\sqrt2}=\exp(-\phi)\mu_1$$
$${\nu_2}'=\nu_2,\quad{\nu_3}'=\nu_3,\quad\cdots$$
$${\mu_2}'=\mu_2,\quad{\mu_3}'=\mu_3,\quad\cdots$$
The vector $\nu_1$ is multiplied by some positive number $e^\phi$, and the covector $\mu_1$ is divided by the same number. The dot product is still ${\mu_1}'\cdot{\nu_1}'=1$.
You can get a negative multiplier for $\nu_1$ simply by the inversion $\sigma_1\mapsto{^-}\sigma_1,\quad\tau_1\mapsto{^-}\tau_1$; this will also negate $\mu_1$. The result is that you can multiply $\nu_1$ by any non-zero Real number, and $\mu_1$ will be divided by the same number.
Of course, this only varies one basis vector in one direction. You could try to rotate the vectors, but a simple rotation in a $\sigma_i\sigma_j$ plane will mix $V$ and $V^*$ together. This problem is solved by an isoclinic rotation in $\sigma_i\sigma_j$ and $\tau_i\tau_j$, which causes the same rotation in $\nu_i\nu_j$ and $\mu_i\mu_j$ (while keeping them separate).
Combine these stretches, reflections, and rotations, and you can generate any invertible linear transformation on $V$, all while maintaining the degeneracy ${\nu_i}^2=0$ and the duality $\mu_i\cdot\nu_j=\delta_{i,j}$. This shows that $V$ and $V^*$ do have the correct "variance".
See also Hestenes' Tutorial, page 5 ("Quadratic forms vs contractions").
Best Answer
$ \newcommand\Cl{\mathrm{Cl}} \newcommand\Ext{{\textstyle\bigwedge}} \newcommand\rev\widetilde \newcommand\rintr{\mathbin{{\llcorner}}} \newcommand\R{\mathbb R} \newcommand\form[1]{\langle#1\rangle} \newcommand\dd{\mathrm{d}} \newcommand\lcontr{\mathbin{{\rfloor}}} \newcommand\rcontr{\mathbin{{\lfloor}}} \newcommand\doub{\mathfrak} \newcommand\doubnabla{\boldsymbol\nabla} \newcommand\adj{\overline} $
Let $Q$ be a quadratic form over an $n$-dimensional vector space $V$ (which for this discussion we will assume is real though this isn't strictly necessary.) Quadratic forms $Q$ and symmetric bilinear forms $B$ are in one-to-one correspondence via $$ B(v, w) = \frac12(Q(v + w) - Q(v) - Q(w)),\quad Q(v) = B(v, v). $$ In this context, metric is another word for symmetric bilinear form.
The Clifford algebra $\Cl(V, Q)$ is essentially the associative algebra generated by $V$ with the relations $v^2 = Q(v)$ for all $v \in V$. This can be formalized by defining $\Cl(V, Q)$ as the algebra with a certain universal property, or by quotienting the tensor algebra by the previously mentioned relations. It is not defined in terms of a wedge product; I would consider such a definition a bad definition.
But there are two important relationships between Clifford algebras and the exterior algebra $\Ext V$. First, $\Cl(V, 0)$ is exactly $\Ext V$, where $0$ is the trivial quadratic form. Second, there is a canonical linear isomorphism (not an algebra isomorphism) $\Cl(V, Q) \cong \Ext V$ for any $Q$. This means that we can view $\Cl(V, Q)$ as $\Ext V$ endowed with a second product, the Clifford product, and one way of constructing $\Cl(V, Q)$ is by defining such a product on $\Ext V$ (but I would never take this as a definition).
In light of this, we can view various operations on $\Ext V$ as operations on $\Cl(V, Q)$. This is very natural with e.g. the Hodge star $\star$. Denote by $\Cl^k(V, Q)$ the linear subspace of $\Cl(V, Q)$ corresponding to $\Ext^k V$. There is an antiautomorphism $A \mapsto \rev A$ of $\Cl(V, Q)$ which reverses all products and is the identity on $V$. We get the Hodge star when $Q$ is non-degenerate and when we choose $I \in \Cl^n(V, Q)$ such that $I^2 = \pm1$ (such elements always square to scalars); this is equivalent to choosing an orientation on $V$. Then the Hodge star is exactly $$ \star A = \rev AI $$ for any $A \in \Cl(V, Q) \cong \Ext V$.
A differential form assigns an element of $(\Ext V)^*$ to each tangent space of a manifold (where we assume the tangent spaces are isomorphic to $V$). There is a canonical isomorphism $(\Ext V)^* \cong \Ext V^*$ induced by a natural bilinear pairing $\form{{-}, {-}} : \Ext V^* \times \Ext V \to \R$, so we usually identify a differential form with an element of $\Ext V^*$. Then the action of a differential form $A^* \in \Ext V^*$ on $B \in \Ext V$ is given by $\form{A^*, B}$, or equivalently the action on vectors $v_1,\dotsc, v_k \in V$ is given by $$ \form{A^*, v_1\wedge\cdots\wedge v_k}. $$ But in the presence of a non-degenerate metric there is an isomorphism $V \cong V^*$, and this extends naturally to algebra isomorphisms $\Ext V \cong \Ext V^*$ and $\Cl(V, Q) \cong \Cl(V^*, Q^*)$ (where $Q^*$ is $Q$ applied to $V^*$ via the isomorphism). This means we can identify differential forms with elements of $\Ext V$. The pairing $\form{{-}, {-}}$ becomes a bilinear form on $\Ext V$, and is (almost) exactly the scalar part of the Clifford product: $$ \form{A, B} = \form{\rev AB}_0,\quad A, B \in \Cl(V, Q) \cong \Ext V. $$ This scalar part is defined either by taking advantage of the grading of $\Ext V$, or as intrinsic to the Clifford algebra via the normalized trace: $$ \form{A}_0 = \frac1{2^n}\mathrm{Tr}(B \mapsto AB),\quad B \in \Cl(V, Q). $$ The reversal $\rev A$ in the above and in the Hodge star is an artifact of how $\form{{-},{-}}$ is defined, and can be done away with if desired. For simplicity, we will use this convention and define $$ \form{A, B} = \form{AB}_0,\quad \star A = AI. $$
In this way, we can represent differential forms as elements of $\Cl(V, Q)$ and use all the tools that comes with that.
As a particular example, we get the exterior derivative as $$ [\dd\form{A_x, {-}}](B) = \form{(\nabla\wedge A_x)B}_0 $$ and so the codifferential as $$ [{\star^{-1}}\dd{\star}\form{A_x, {-}}](B) = \form{(\nabla\wedge(A_xI))I^{-1}B}_0 = \form{(\nabla\lcontr A_x)B}_0 $$ where $\lcontr$ is the left contraction $$ A\lcontr B = \form{AB}_{k-j},\quad A \in \Cl^j(V, Q),\quad B \in \Cl^k(V, Q), $$ i.e. the projection onto the exterior algebra grade ${k-j}$ component, which we define to be zero if $j > k$. This is the manifestation of the interior product in the presence of a metric, as is the closely related right contraction $\rcontr$. To summarize, the exterior derivative and codifferentials are expressed in $\Cl(V, Q)$ as $$ \nabla\wedge A_x,\quad \nabla\lcontr A_x. $$ But for any $v \in V$ and $A \in \Cl(V, Q)$ $$ vA = v\wedge A + v\lcontr A $$ so we see that the sum of the exterior derivative and codifferential is the geometric derivative $$ \nabla A_x = \nabla\wedge A + \nabla\lcontr A $$ where on the left we are using the Clifford product. This is extremely useful; for example, Maxwell's equations become exactly one equation $\nabla F = J$, and (at least in flat space) $\nabla$ has a Green's function which allows us to directly solve for $F$ in terms of $J$.
This is the beginning of geometric calculus; I would recommend chapter 6 of Doran and Lasenby's Geometric Algebra for Physicists (2003) for more. (Really, the whole book is worth a read.) In brief, this approach to differential geometry (of course with a focus on Riemannian geometry) involves embedding our manifold in flat space, where then one of the major objects of study becomes the bivector-valued shape tensor $S_x$ which encodes much of the extrinsic and intrinsic geoemtry of the manifold. For instance, the covariant derivative (i.e. Levi-Civita connection) can be expressed as $$ DA_x = \partial A_x + \partial_a(S_x(a)\times A_x) $$ where $X\times Y = \tfrac12(XY - YX)$, the tangential derivative $\dot\partial = P_x(\dot\nabla)$ is the "projection" of $\nabla$ onto the tangent space of the manifold, and $\partial_a$ is the tangential derivative with respect to $a$ (actually representing a kind of tensor contraction since $S_x$ is linear).
The introduction of a metric obviously takes us away from pure differential forms. We can actually try to do away with the metric: the space $\doub W = V^*\oplus V$ has a natural bilinear form $$ \form{v^* + v, w^* + w} = v^*(w) + w^*(v) $$ with quadratic form $$ \form{v^* + v, v^* + v} = 2v^*(v). $$ We then consider the Clifford algebra $\Cl(\doub W)$. Since $\form{v^*, v^*} = \form{v, v} = 0$, the subalgebras generated by $V^*$ and $V$ are exactly $\Ext V^*$ and $\Ext V$. Not only that, but if $M$ is a manifold such that $T_xM \cong V$, then the tangent spaces of the cotangent bundle are naturally isomorphic to $\doub W$. This makes $\Cl(\doub W)$ a potentially very natural setting in which to study differential forms.
However, I have yet to study this construct to my satisfaction, so I will stop here.