What string theory aims to do is to unify GR with quantum field theory, not quantum mechanics. Before discussing what string theory does, we need to correct some misconceptions you have.
I’m not sure what you mean by time “varying” vs “being a constant”; I assume you mean that in quantum mechanics time is a parameter and space is an operator, and we don’t have Lorentz invariance. However, that comes up when we compare QM with special relativity too, and led to the development of quantum field theory, which fully incorporates special relativity.
Quantum mechanics solves for the behaviour of a system composed of a certain number of particles by finding the eigenstates of a Hamiltonian operator. Quantum field theory does something completely different. Why? Because quantum mechanics quantises classical mechanics, whereas classical field theory quantises classical field theory. Forget everything you know about quantum theory for the moment and just compare these two equations: $$\ddot{\mathbf{x}}=-\boldsymbol{\nabla}V,\, \partial_\mu F^{\mu\nu}=\mu_0 j^\nu.$$Both are equations of motion. The first can be obtained as an Euler-Lagrange equation of the action $$S=\int dt\left( \frac{1}{2}m\dot{x}^2-V\left( x\right) \right),$$while the second can be obtained as one of several Euler-Lagrange equations of the action $$S=\int d^4x \left(-\frac{1}{4\mu_0}F_{\mu\nu}F^{\mu\nu}-A_\nu j^\nu\right)$$where we abbreviate $d^4x=dtd^3x,\,d^3x=dtdxdydz$. Just as classical mechanics writes an action in terms of functions of time called coordinates and time derivatives thereof (and possibly also time itself, in which case energy is not conserved), so classical field theory writes an action in terms of functions of spacetime called fields and spacetime derivatives thereof (and possibly also spacetime itself, in which case energy is not conserved if in particular there is an explicit time dependence).
What does all this have to do with the Hamiltonian being ditched? Well, when we turn quantisation back on again we discover we can no longer have a probability amplitude for one particle's location. 669
If you rearrange the time-dependent <PERSON> equation as $\dot{\psi}=i\left( \frac{\hbar}{2m}\nabla^2-\frac{V}{\hbar} \right)\psi$, you can use $\rho=\psi^\ast \psi$ to prove that $\dot{\rho}+\boldsymbol{\nabla}\cdot\mathbf{j}=0$ for probability 3-current $\mathbf{j}=\frac{i\hbar}{2m}\left(\psi\boldsymbol{\nabla}\psi^{\ast}-\psi^{\ast}\boldsymbol{\nabla}\psi\right)$. If this probability interpretation can survive in relativity, we need $\partial_\mu j^\mu = 0$ with $\int d^3 x j^0 = 1$ for some $j^\mu$ expressible in terms of solutions of a relativistic variant of the <PERSON> equation. But in theory, the <PERSON> equation can be interpreted as an equation in some field $\psi$ that has no probabilistic interpretation. Indeed, any relativistic upgrade ends up with solutions for which $\int d^3 x j^0 \le 0$. The ultimate resolution is to see $j^0$ as a difference bewteen particle and antiparticle densities and take the fields in the theory as descriptions of the entire population of such particles and antiparticles in the universe (e.g. the Dirac spinor describes all electrons and positrons). But if particles are now measurable properties of fields in the same way we're used to thinking of momentum as a measurable property of one particle, it is the Lagrangian, not the Hamiltonian, that takes centre stage. The theory doesn't need energy conservation to work, but it gets it anyway in <PERSON> space (or, indeed, any spacetime whose metric tensor's determinant is time-independent).
Now let's talk about renormalisation. All current field theories are low-energy variants, obtainable as discussed here, of as yet unknown high-energy theories. 976