I) In the Lagrangian $L(q(t),\dot{q}(t),t)$, one must distinguish between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.$^1$
However, the implicit time dependence in the Lagrangian $L$ does only make sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} [t_i,t_f]~\stackrel{q}{\longrightarrow}~\mathbb{R}^n.$$ The implicit time dependence would typically be different for another path.
II) In fact a (possibly virtual) path (1) is technically speaking not the input for a Lagrangian $L$. Rather the Lagrangian $$\tag{2} \mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\stackrel{L}{\longrightarrow}~\mathbb{R}$$ is a function
$$\tag{3} (q,v,t)~\mapsto~ L(q,v,t) $$
(as opposed to a functional) that only depends on
an instant $t\in[t_i,t_f]$,
an instantaneous position$^2$ $q\in\mathbb{R}^n$, and
an instantaneous velocity $v\in\mathbb{R}^n$;
not the past, nor the future.
Notice that we here use the symbol $v$ rather than the notation $\dot{q}\equiv \frac{dq}{dt}$. This is because the ability to differentiate $\frac{dq}{dt}$ would imply that we know (at least a segment of) a path (1) rather than just information about an instantaneous state $(q,v,t)$ of the system.
III) In contrast, the action
$$\tag{4} S[q] ~:=~ \int_{t_i}^{t_f}dt \ L(q(t),\dot{q}(t),t)$$
is a functional (as opposed to a function) that depends on a (possibly virtual) path (1).
For more details, such as, e.g., an explanation how calculus of variations works, why $q$ and $v$ are independent variables in the Lagrangian (3) but dependent variables in the action (4), etc.; see e.g. this related Phys.SE post and links therein.
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$^1$ By the way, if the Lagrangian $L(q,v)$ has no explicit time dependence, then the energy
$$ \tag{5} h(q,v)~:=~v^i \frac{\partial L(q,v)}{\partial v^i}-L(q,v) $$
is conserved, cf. Noether's theorem.
$^2$ Here $q\in\mathbb{R}^n$ denotes an $n$-tuple, as opposed to eq. (1) where $q$ denotes a path/curve.
