The answer is yes, the essence of Noether's theorem for linear and angular momentum can be understood without using the Lagrangian (or Hamiltonian) formulation, at least if we're willing to focus on models in which the equations of motion have the form
$$
m_n\mathbf{\ddot x}_n = \mathbf{F}_n(\mathbf{x}_1,\mathbf{x}_2,...)
\tag{1}
$$
where $m_n$ and $\mathbf{x}_n$ are the mass and location of the $n$-th object, overhead dots denote time-derivatives, and $\mathbf{F}_n$ is the force on the $n$-th object, which depends on the locations of all of the objects.
(This answer still uses math, but it doesn't use Lagrangians or Hamiltonians. An answer that doesn't use math is also possible, but it would be wordier and less convincing.)
The inputs to Noether's theorem are the action principle together with a (continuous) symmetry. For a system like (1), the action principle can be expressed like this:
$$
\mathbf{F}_n(\mathbf{x}_1,\mathbf{x}_2,...)
= -\nabla_n V(\mathbf{x}_1,\mathbf{x}_2,...).
\tag{2}
$$
The key point of this equation is that the forces are all derived from the same function $V$.
Loosely translated, this says that if the force on object $A$ depends on the location of object $B$, then the force on object $B$ must also depend (in a special way) on the location of object $A$.
First consider linear momentum. Suppose that the model is invariant under translations in space. In the context of Noether's theorem, this is a statement about the function $V$. This is important! If we merely assume that the system of equations (1) is invariant under translations in space, then conservation of momentum would not be implied. (To see this, consider a system with only one object subject to a location-independent force.) What we need to do is assume that $V$ is invariant under translations in space. This means
$$
V(\mathbf{x}_1+\mathbf{c},\mathbf{x}_2+\mathbf{c},...) =
V(\mathbf{x}_1,\mathbf{x}_2,...)
\tag{3}
$$
for any $\mathbf{c}$. The same condition may also be expressed like this:
$$
\frac{\partial}{\partial\mathbf{c}}V(\mathbf{x}_1+\mathbf{c},\mathbf{x}_2+\mathbf{c},...) = 0,
\tag{4}
$$
where $\partial/\partial\mathbf{c}$ denotes the gradient with respect to $\mathbf{c}$. Equation (4), in turn, may also be written like this:
$$
\sum_n\nabla_n V(\mathbf{x}_1\,\mathbf{x}_2,\,...) = 0.
\tag{5}
$$
Combine equations (1), (2), and (5) to get
$$
\sum_n m_n\mathbf{\ddot x}_n = 0,
\tag{6}
$$
which can also be written
$$
\frac{d}{dt}\sum_n m_n\mathbf{\dot x}_n = 0.
$$
This is conservation of (total) linear momentum.
Now consider angular momentum. For this, we need to assume that $V$ is invariant under rotations. To be specific, assume that $V$ is invariant under rotations about the origin; this will lead to conservation of angular momentum about the origin.
The analogue of equation (5) is
$$
\sum_n\mathbf{x}_n\wedge \nabla_n V(\mathbf{x}_1\,\mathbf{x}_2,\,...) = 0
\tag{7}
$$
where the components of $\mathbf{x}\wedge\nabla$ are $x_j\nabla_k-x_k\nabla_j$. (For three-dimensional space, this is usually expressed using the "cross product", but I prefer a formulation that works in
any number of dimensions so that it can be applied without hesitation to easier cases like two-dimensional space.) Equation (7) expresses the assumption that $V$ is invariant under rotations about the origin. As before, combine equations (1), (2), and (7) to get
$$
\sum_n \mathbf{x}_n\wedge m_n\mathbf{\ddot x}_n = 0,
\tag{8}
$$
and use the trivial identity
$$
\mathbf{\dot x}_n\wedge \mathbf{\dot x}_n = 0
\tag{9}
$$
(because $\mathbf{a}\wedge\mathbf{b}$ has components $a_jb_k-a_kb_j$)
to see that equation (8) can also be written
$$
\frac{d}{dt}\sum_n \mathbf{x}_n\wedge m_n\mathbf{\dot x}_n = 0.
\tag{10}
$$
This is conservation of (total) angular momentum about the origin.
Best Answer
Try and think of a Noether charge besides a momenta (linear, angular, Hamiltonian). It's rather hard to do in point particle mechanics because there really aren't any we talk about (it's easier in field theory to come up with things that aren't just momentum).
So, let me pose the following question, which I think will make the point I want to get at: without looking at your Lagrangian, can you tell me what the correct expression for the momentum is? If you say $mv$, that's not always correct. For example, if your point particle charged and in the presence of an electromagnetic field the standard momentum ends up being something like $m\boldsymbol v-q\boldsymbol v\cdot\boldsymbol A/c$ where $\boldsymbol A$ is the vector potential and $c$ is the speed of light.
So the point here is that the actual expression for momentum depends entirely on the Lagrangian you're looking at. In this view, the word "momentum" is nothing more than indicating the Noether charge associated to translation invariance. I could very easily come up with a different transform and then give its associated Noether charge a special name, but at the end of the day, it's nothing more than "the Noether charge associated to the given symmetry."
As a result, the statement "momentum is the Noether charge associated to translations" is of a different nature than "what symmetry would give rise to acceleration conservation." The latter question is more like "what symmetry gives rise to the conservation of $mv$." The answer is going to depend on the details of the Lagrangian you have on hand. For many standard Lagrangians, the answer to the question "what symmetry gives rise to the conservation of $mv$" will be "translations," but as I've pointed out above, if our Lagrangian is the one which gives us the Lorentz force equations, $mv$ is not conserved and even if it was, it wouldn't be the momentum because it differs from the actual Noether charge by an addition of the vector potential.
So when you ask "what symmetry would give rise to acceleration conservation," really you're asking a question that requires a specific Lagrangian in order to answer.