I consider here only the classical side, where one can directly compare the Lagrangian and the Hamiltonian formulation. Here the two notions of symmetry are equivalent under suitable hypotheses as I explain below.
The passage from classical Hamiltonian formulation to Quantum Mechanics is much more delicate but, roughly speaking, we can say that classical (continuous) symmetries are preserved at quantum level when a "quantization procedure" is known.
Regarding Lagrangian symmetries vs Hamiltonan ones at classical level, I can say what follows.
Every one-parameter group of continuous (smooth) geometric transformation $$q'^k=q'^k(\epsilon, t, q^1,q^2,\ldots, q^n)\quad for \quad k=1,\ldots, n\tag{1}$$
so that, accordingly,
$$\dot{q}'^k = \frac{\partial q'^k}{\partial t}+\sum_j \frac{\partial q'^k}{\partial q^j} \dot{q}^j\:,\quad for \quad k=1,\ldots, n.$$
Above $\epsilon \in \mathbb{R}$ is the parameter of the transformation, so that
$$q'(0,t, q)= q\quad and\quad q'(\epsilon,t, q'_{\epsilon'}(t,q))= q'(\epsilon+\epsilon',t, q)$$
can be implemented (at least locally) through a new system of coordinates $Q^1,\ldots, Q^n$ where the parameter $\epsilon$ becomes a coordinate $Q^1=\epsilon$. Within this setup (1) simply reads
$$Q'^1= Q^1+\epsilon,\quad Q'^2 = Q^2, \quad \ldots \quad Q'^n=Q^n \:.$$
At this point there is a general important identity that is consequence of the Legendre transformation
$$\frac{\partial H(t,q,p)}{\partial q^k} = - \frac{\partial L(t,q,\dot{q})}{\partial q^k}\tag{2}$$
This is valid independently of the choice of the coordinates when assuming that they are related through the Legendre transformation
$$t=t, \quad q^k=q^k, \quad p_k = \frac{\partial L(t,q,\dot{q})}{\partial \dot{q}^k}$$
Specializing (2) to the coordinates $Q^1,\ldots, Q^n$ adapted to the considered group of symmetries we have that, if $L$ is invariant under the one-parameter group of symmetries (1), then
$$\frac{\partial H(t,Q,P)}{\partial Q^1} = - \frac{\partial L(t,Q,\dot{Q})}{\partial Q^1}=0\tag{3}\:.$$
Coming back to the original coordinates
$$\frac{\partial H(t,q'(\epsilon,t, q),p'(\epsilon,t, q))}{\partial \epsilon} =\frac{\partial H(t,Q,P)}{\partial Q^1} =0\:.$$
This proves that invariance of the Lagrangian under a group of transformation (1) is equivalent to invariance of the Hamiltonian under the same group of transformations.
The crucial point is that the symmetry does not involve the dot coordinates $\dot{q}$ inthe right-hand side of (1). If this condition is not valid then the correspondence has to be changed. The Hamiltonian formulation is easier to handle than the Lagrangian one in view of the canonical transformation formalism.
