Ok, let us start from scratch. A function $g: \mathbb R^n \to \mathbb R$ with $f \in C^2(\mathbb R^n)$ is said to be convex if its Hessian matrix (i.e. the one with coefficients $\partial^2 f/\partial x_i \partial x_j$) is everywhere (strictly) positively defined.
Let $\Omega \subset \mathbb R \times \mathbb R^n$ be an open set and focus on a jointly $C^2$ Lagrangian function $\Omega \times \mathbb R^n \ni (t,q,\dot{q}) \mapsto L(t, q, \dot{q}) \in \mathbb R$.
For fixed $(t,q) \in \Omega$, $L$ is assumed to be convex as a function of $\dot{q}$. In other words $\mathbb R^n \ni \dot{q} \mapsto L(t, q, \dot{q}) \in \mathbb R$ is supposed to be convex.
Referring to either systems made of material points or solid bodies, convexity arises form the structure of the kinetic energy part of Lagrangians, which are always of the form $T(t, q, \dot{q}) - V(t, q)$, even considering generalized potentials $V(t,q, \dot{q})$ with linear dependence on $\dot{q}$, as is the case for inertial or electromagnetic forces or inertial forces also in the presence of holonomic ideal constraints.
The associated Hamiltonian function is defined as the Legendre transformation of $L$ with respect to the variables $\dot{q}$. In other words:
$$H(t,q,p) := \max_{\dot{q} \in \mathbb R^n}\left[p\cdot \dot{q} - L(t, q, \dot{q})\right]\qquad (1)$$
Within our hypotheses on $L$, from the general theory of Legendre transformation, it arises that, for fixed $(t,q) \in \Omega$, a given $p \in \mathbb R^n$ is associated with exactly one $\dot{q}(p)_{t,q} \in \mathbb R^n$ where the maximum of the RHS in (1) is attained (for $n=1$ the proof is quite evident, it is not for $n>1$).
Since $\dot{q}(p)_{t,q} $ trivially belongs to the interior of the domain of the function $\mathbb R^n \ni \dot{q} \mapsto p\cdot \dot{q} - L(t, q, \dot{q})$, it must be:
$$\left.\nabla_{\dot{q}} \right|_{\dot{q}= \dot{q}(p)_{t,q}} \left( p\cdot \dot{q} - L(t, q, \dot{q})\right) =0\:.$$
In other words (always for fixed $t,q$):
$$p = \left.\nabla_{\dot{q}} \right|_{\dot{q}(p)_{t,q}} L(t, q, \dot{q})\:, \quad \forall \dot{q} \in \mathbb R^n\qquad (2)$$
As a consequence, (always for fixed $(t,q)\in \Omega$) the map $\mathbb R^n \ni p \mapsto \dot{q}(p)_{t,q} \in \mathbb R^n$
is injective, because it admits a right inverse given by the map
$\mathbb R^n \ni \dot{q} \mapsto \nabla_{\dot{q}} L(t, q, \dot{q})$ which, in turn, is surjective. However the latter map is also injective, as one easily proves
using the convexity condition and the fact that the domain $\mathbb R^n$ is trivially convex too. The fact that the $\dot{q}$-Hessian matrix of $L$ is non-singular also implies that the map (2) is $C^1$ with its inverse.
Summing up, the map (2) is a $C^1$ diffeomorphism from $\mathbb R^n$ onto $\mathbb R^n$ and, from (1), we have the popular identity describing the interplay of the Hamiltonian and Lagrangian functions as:
$$H(t,q,p) = p\cdot \dot{q} - L(t, q, \dot{q})\qquad (3)$$
which holds true when $p \in \mathbb R^n$ and $\dot{q} \in \mathbb R^n$ are related by means of the $C^1$ diffeomorphism from $\mathbb R^n$ onto $\mathbb R^n$ (for fixed $(t,q)\in \Omega$):
$$p = \nabla_{\dot{q}} L(t, q, \dot{q})\:, \quad \forall \dot{q} \in \mathbb R^n\qquad (4)\:.$$
By construction, $H= H(t,q,p)$ is a jointly $C^1$ function defined on $\Gamma := \Omega \times \mathbb R^n$. I stress that $L$ is defined on the same domain $\Gamma$ in $\mathbb R^{2n+1}$. The open set $\Gamma$ is equipped by the diffeomorphism:
$$\psi: \Gamma \ni (t,q, \dot{q}) \mapsto (t,q, p) \in \Gamma \qquad (4)'$$
where (4) holds.
Let us study the relationship between the various derivatives of $H$ and $L$.
I remark that I will not make use of Euler-Lagrange or Hamilton equations anywhere in the following.
Consider a $C^1$ curve $\gamma: (a,b) \ni t \mapsto (t, q(t), \dot{q}(t)) \in \Gamma$, where $t$ has no particular meaning and $\dot{q}(t)\neq \frac{dq}{dt}$ generally. The diffeomorphism $\psi$ transform that curve into a similar $C^1$ curve $t \mapsto \psi(\gamma(t)) = \gamma'(t)$ I will also indicate by $\gamma': (a,b) \ni t \mapsto (t, q(t), p(t)) \in \Gamma$.
We can now evaluate $H$ over $\gamma'$ and $L$ over $\gamma$ and compute the total temporal derivative taking (3) and (4) into account, i.e. we compute:
$$\frac{d}{dt} H(t, q(t),p(t)) = \frac{d}{dt}\left(p(t) \dot{q}(t) - L(t,q(t),p(t)) \right)\:.$$
Computations gives rise almost immediately to the identity, where both sides are evaluated on the respective curve:
$$\frac{\partial H}{\partial t} + \frac{dq}{dt}\cdot \nabla_q H
+ \frac{dp}{dt}\cdot \nabla_p H = \frac{dp}{dt}\dot{q} + p \frac{d\dot{q}}{dt} -\frac{\partial L}{\partial t} - \frac{dq}{dt}\cdot \nabla_q L
- \frac{d\dot{q}}{dt}\cdot \nabla_{\dot{q}} L \:.$$
In the RHS, the second and the last term cancel each other in view of (4), so that:
$$\frac{\partial H}{\partial t} + \frac{dq}{dt}\cdot \nabla_q H
+ \frac{dp}{dt}\cdot \nabla_p H = \frac{dp}{dt}\dot{q} -\frac{\partial L}{\partial t} - \frac{dq}{dt}\cdot \nabla_q L \:.$$
Rearranging the various terms into a more useful structure:
$$\left(\frac{\partial H}{\partial t}|_{\gamma'(t)} + \frac{\partial L}{\partial t}|_{\gamma(t)}\right) +
\frac{dq}{dt}\cdot \left( \nabla_q H|_{\gamma'(t)} + \nabla_q L|_{\gamma(t)}\right) +
\frac{dp}{dt}\cdot \left(\nabla_p H|_{\gamma'(t)} - \dot{q}|_{\gamma(t)}\right) =0\:.\qquad (5)$$
Now observe that actually, since $\gamma$ is generic, $\gamma(t)$ and $\gamma'(t)= \psi(\gamma(t))$ are generic points in $\Gamma$ (however connected by the transformation (4)). Moreover, given the point $(t,q, \dot{q}) = \gamma(t) \in \Gamma$, we are free to choose the derivatives $\frac{dq}{dt}$ and (using the diffeomorphism) $\frac{dp}{dt}$ as we want, fixing $\gamma$ suitably. If we fix to zero all these derivatives, (5) proves that, if $(t,q, \dot{q})$ and $(t,q,p)$ are related by means of (4):
$$\left(\frac{\partial H}{\partial t}|_{(t,q,p)} + \frac{\partial L}{\partial t}|_{(t,q, \dot{q})}\right) =0\:.$$
This result does not depend on derivatives $dq/dt$ and $dp/dt$ since they do not appear as arguments of the involved functions. So this result holds everywhere in $\Gamma$ because $(t,q, \dot{q})$ is a generic point therein.
We conclude that (5) can be re-written as:
$$\frac{dq}{dt}\cdot \left( \nabla_q H|_{\gamma'(t)} + \nabla_q L|_{\gamma(t)}\right) +
\frac{dp}{dt}\cdot \left(\nabla_p H|_{\gamma'(t)} - \dot{q}|_{\gamma(t)}\right) =0\:.\qquad (5)'$$
where again, we are considering a generic curve $\gamma$ as before.
Fixing such curve such that all components of $\frac{dq}{dt}$ and $\frac{dp}{dt}$ vanish except for one of them, for instance $\frac{dq^1}{dt}$, we find:
$$\left(\frac{\partial H}{\partial q^1}|_{(t,q,p)} + \frac{\partial L}{\partial q^1}|_{(t,q, \dot{q})}\right) =0\:,$$
if $(t,q, \dot{q})$ and $(t,q,p)$ are related by means of (4), and so on.
Eventually we end up with the following identities, valid when $(t,q, \dot{q})$ and $(t,q,p)$ are related by means of (4)
$$\frac{\partial H}{\partial t}|_{(t,q,p)} =- \frac{\partial L}{\partial t}|_{(t,q, \dot{q})}\:, \quad \frac{\partial H}{\partial q^k}|_{(t,q,p)} =- \frac{\partial L}{\partial q^k}|_{(t,q, \dot{q})}\:, \quad
\frac{\partial H}{\partial p_k}|_{(t,q,p)} = \dot{q}^k\:.
\quad (6)$$
The last identity is the one you asked for. As you see, the found identities rely upon the Legendre transformation only and they do not consider Euler-Lagrangian equations or Hamilton ones.
However, exploiting these identities, it immediately arises that $\gamma$ verifies EL equations:
$$\frac{d}{dt} \frac{\partial L}{\partial \dot{q}^k} - \frac{\partial L}{\partial q^k}=0\:,\quad \frac{dq^k}{dt} = \dot{q}^k\quad k=1,\ldots, n$$
if and only if the transformed curve
$\gamma'(t) := \psi(\gamma(t))$ verifies Hamilton equations.
$$\frac{d p_k}{dt} = -\frac{\partial H}{\partial q^k} \:, \quad \frac{dq^k}{dt} = \frac{\partial H}{\partial p_k}\quad k=1,\ldots, n\:.$$
Indeed, starting from a curve $\gamma(t) = (t, q(t), \dot{q}(t))$, the first EL equation, exploiting (4) (which is part of the definition of $\psi$) and the second identity in (6), becomes the first Hamilton equation for the transformed curve $\psi (\gamma(t))$. Moreover, the second EL equation, making use of the last identity in (6), becomes the second Hamilton equation for the transformed curve. This procedure is trivially reversible, so that, starting from Hamilton equations, you can go back to EL equations.
The first identity in (6) it not used here. However it implies that the system is or is not invariant under time translations simultaneously in Lagrangian and Hamiltonian formulation (in both cases, that invariance property implies the existence of a constant of motion which is nothing but $H$ represented with the corresponding variables either Lagrangian or Hamiltonian).
As a final comment notice that (3) and the last identity in (6) (which is nothing but the inverse function of (2) at fixed $(t,q)$) imply
$$L(t, q, \dot{q}) = \nabla_p H(t,q,p) \cdot p - H(t,q,p)\:,$$
where (2) is assumed to connect Lagrangian and Hamiltonian variables.
Best Answer
The Lagrangian and Hamiltonian formalisms are exactly equivalent, so any physical observable that can be computed in one formalism, can be computed in the other, and the results must match.
You gave the argument for this statement yourself in your question. You can derive the Hamiltonian formalism from the Lagrangian formalism, and vice versa, so they are equivalent.
Based on some clarifying discussion in the comments, I will add that I think what this question is getting at is whether there are different interpretations of classical mechanics, similar to different interpretations of quantum mechanics (including both what is normally meant by interpretations of quantum mechanics, but also other issues such as the fact that you can look at quantum field theory as a theory of quantized fields or a theory of an indefinite number of identical particles). I would tend to say that classical mechanics is much better understood than quantum field theory, both because it is a more mature subject that has been around longer, and because it is intrinsically less conceptually complicated, and therefore there are not major differences in interpretation of the meaning of classical physics. At the end of the day, a classical theory computes a particle trajectory, or the evolution of a field, as a solution to a differential equations. The different formalisms give different mathematical representations, but at the end of the day typically it is clear what we mean by a particle's trajectory.