"how can you be sure there does not exist a Lagrangian depending explicitly on time such that the partial derivative with respect to time vanishes for all "
If the question is that, it does not make much sense. That is because "depending explicitly on time" exactly means that $\partial L/\partial t \neq 0$ somewhere.
So I assume that the question is whether or not the two conditions
$$\partial L/\partial t = 0\quad\text{everywhere}$$
and
$$\frac{dH}{dt}=0\quad\text{on every solution of the EL equations}$$
are really equivalent. It is known that the former implies the latter, but the converse implication is not so evident.
Above and henceforth,
$$H(t,q,\dot{q}) := -L(t,q,\dot{q})+ \sum_{k=1}^n \dot{q}^k
\frac{\partial L}{\partial \dot{q}^k}.$$
Under natural hypotheses in Mechanics, the two conditions above are actually equivalent though of very different nature as $\partial L/\partial t=0$ does not know anything about the solutions of the EL equations.
If (a) the Lagrangian has standard form $$L= \sum_{h,k} a_{hk}(t,q) \dot{q}^h \dot{q}^k + \sum_{k} b_{k}(t,q) \dot{q}^k+ c(t,q)\:,\tag{1}$$ (b) it is jointly $C^2$, and (c) $$\det\left[\frac{\partial^2 L}{\partial \dot{q}^h \partial \dot{q}^k}\right] = \det [a_{hk}(t,q)]\neq 0\quad\text{everywhere}$$
(this requirement does not depend on the used coordinate patch of type $t,q,\dot{q}$)
then the Euler-Lagrange equations satisfy the a local existence and uniqueness theorem of their solutions:
THEOREM 1. With the above conditions (a),(b),(c), given $(t_0,q_0,\dot{q}_0)$ there is a unique solution
$$\gamma(t) = (t,q(t),\dot{q}(t)), \quad t \in I \ni t_0$$
of the E.L. equations
$$\dot{q}^k(t) = \frac{dq^k}{dt}|_{\gamma(t)}\:, \quad \frac{d}{dt} \left(\frac{\partial L}{\partial \dot{q}^k}|_{\gamma(t)}\right)- \frac{\partial L}{\partial q^k}|_{\gamma(t)}=0, \quad k=1,\ldots, n$$
in a sufficiently small neighborhood of $(t_0,q_0,\dot{q}_0)$ with these initial conditions: $(t_0,q(t_0),\dot{q}(t_0))= (t_0,q_0,\dot{q}_0)$.
(Actually this theorem holds globally, but here the local version is sufficient.)
This theorem has the immediate corollary that answers the question:
COROLLARY. With the hypotheses of theorem 1, in every coordinate patch with coordinate $t,q,\dot{q}$, these two facts are equivalent:
$$\frac{\partial L}{\partial t}=0 \quad\text{everywhere in the said coordinate patch}\tag{A}$$
$$\frac{dH(\gamma(t))}{dt}=0\quad\text{for every (local) EL solution }\gamma=\gamma(t)\text{ in the said coordinate patch.}\tag{B}$$
PROOF. We know that on every solution $\gamma$
$$\frac{dH(\gamma(t))}{dt} = - \frac{\partial L}{\partial t}|_{\gamma(t)}\tag{2}\:.$$
This immediately yields (A) $\Rightarrow$ (B). Now suppose that (B) holds true. Consider a kinetic state $(t_0,q_0,\dot{q}_0)$ in the considered coordinate patch. There is a solution $\gamma$ satisfying $\gamma(t_0)= (t_0,q_0,\dot{q}_0)$. Due to (B) and (2) we also have
$\frac{\partial L}{\partial t}|_{t_0,q_0,\dot{q}_0}=0$. Since $(t_0,q_0,\dot{q}_0)$ is generic, we have that (B) $\Rightarrow$ (A). QED
Reference
