Comments to the question (v1):
I) One must distinguish in the Lagrangian $L(q(t),\dot{q}(t),t)$ between implicit time dependence via the variables $q(t)$ and $\dot{q}(t)$, and explicit time dependence.
However, the implicit time dependence in the Lagrangian $L$ only makes sense in the context of a fixed (but arbitrary, possibly virtual) path $$\tag{1} q:[t_i,t_f]\to \mathbb{R}^n.$$ 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 $$\tag{2} L:\mathbb{R}^n\times\mathbb{R}^n\times [t_i,t_f]~\to~\mathbb{R}$$ 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 $q\in\mathbb{R}^n$, 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 an instantaneous position.
III) In contrast, the action $S$ is a functional (as opposed to a function) that depends on a (possibly virtual) path (1), i.e. both the past and the future.
For more details, and how calculus of variations work, see also e.g. this related Phys.SE post and links therein.