I guess there will not exist a $\tilde \varphi_X \in C^{\infty}(\tilde M, \mathbb{R})$ if $\omega_\epsilon$ is not invariant under $Re(X)$. But if you allow $C^{\infty}(\tilde M, \mathbb{C})$, then the following general fact is true :
$\textbf{Proposition}:$ A holomorphic vector field has a hamiltonian (possibly complex valued) (with respect to any Kahler form) if and only if it vanishes at some point.
This fact is an exercise in "An introduction to extremal Kahler metrics" by Gabor Szekelyhidi. He attributes this fact to LeBrun-Simanca.
One possible way of proving the Proposition is the following - Let $(M,\omega)$ be a Kahler manifold and $X$ a holomorphic vector field such that $X(p)=0$ for some $p\in M$. Since $X$ is holomorphic, $i_X\omega$ is $\bar\partial$-closed $(0,1)$ form, and represents a co-homology class. So there must exist a unique harmonic $(0,1)$ form $\alpha \in [i_X\omega]$. If we can show that $\alpha = 0$, we are done, since then $i_X\omega$ would have to be $\bar \partial$-exact which is equivalent to saying that $X$ is Hamiltonian.
$\textbf{Claim}$ : $\bar \alpha (X) = 0$. To see this, we first notice that since we are working on Kahler manifold and $\alpha$ is $\bar\partial$-harmonic $(0,1)$ form, $\bar\alpha$ is a $\bar\partial$-harmonic $(1,0)$ form. So $\bar\partial \bar\alpha = 0$. But then this implies that $\bar\alpha(X)$ is a holomorphic function which vanishes at $p\in M$, and so must be identically zero. We can now complete the proof -
$$\int_{M}|\alpha|^2\frac{\omega^n}{n!} = \int_{M}\bar\alpha\wedge\alpha\wedge\frac{\omega^{n-1}}{(n-1)!} = \int_{M}\bar\alpha \wedge i_{X}\omega \wedge \frac{\omega^{n-1}}{(n-1)!} = \int_{M}\bar\alpha(X)\frac{\omega^n}{n!}=0.$$
The second equality follows from integration by parts, since $\alpha = i_X\omega + \bar\partial f$ for some function $f$, and $\omega$ and $\bar\alpha$ are $\bar\partial$-closed. Hence $\alpha = 0$, and this completes the proof of the Proposition.
So in your problem since the lifted vector does vanish (on all exceptional divisors), it will have a (possibly complex valued) hamiltonian with respect to any Kahler form on $\tilde M$.
Let's assume that your connection is flat. This is a reasonable assumption, since more generally, you would at least need that your isomorphisms preserve the connection 2-form, which is a pretty stringent (non-topological) requirement in dimensions $\geq 4$, so unless you have a good reason to prescribe a specific curvature form, it's best to stick to the flat setting. Then, you're defining what is known as locally conformal symplectic (LCS) geometry. Much of the literature also takes $L$ to be oriented implicitly, though it's not necessary. There are a few definitions of LCS geometry in the literature, so when you look, you'll have to be careful about translating, but let me at least explain the terminology and provide a hint of part of the translation.
LCS geometry essentially falls out of work of Cartan on pseudo-groups (though historically, it really wasn't taken up until work of Vaisman in the 70's and 80's), the idea being that you can define a geometry in terms of atlases but in which transition maps are allowed to be symplectic (i.e. we're talking about legitimate symplectic forms $\Omega$) up to homothety ($\phi^*\Omega = c\Omega$ for $c \neq 0$ locally constant). You may ask how your definition is legitimately symplectic ($d\Omega = 0$) as opposed to twisted-symplectic ($d_{\nabla}\omega = 0$). The basic translation is as follows. The dual bundle $L^*$ is foliated by the graphs of parallel sections, and each leaf comes with a canonical 2-form, with value at $\phi \in L^*_p$ given by $\Omega_\phi := \langle \phi, \omega \rangle$. Then each leaf, aside from the zero section, is legitimately symplectic, but since we don't have a preferred choice of leaf, we get that we are locally working at all scales.
From this perspective, your connection is really defining a holonomy class in $H^1(M;\mathbb{R}^*)$ which is enough to determine an isomorphism of flat bundles. You can then ask for LCS structures for that given holonomy. You can think that if you take the universal cover $\widetilde{M}$, then it comes with a legitimate symplectic form (up to global scale) such that deck transformations act by homotheties. If you require all homotheties to be positive, so that your holonomy class is in $H^1(M;\mathbb{R}_+) \cong H^1(M;\mathbb{R})$, you're in the setting in which $L$ is oriented.
To answer your questions:
(1) Yes. For the simplest example, if $(Y,\xi = \ker \alpha)$ is a contact manifold (meaning $\alpha \wedge d\alpha^n \neq 0$), then $(\mathbb{R} \times Y, d(e^t\alpha) = e^t(dt \wedge \alpha + d\alpha))$ is symplectic, and the $\mathbb{Z}$-action $t \mapsto t + T$ is a symplectic transformation up to homothety. Hence, the quotient $\mathbb{R}/T\mathbb{Z} \times Y$ comes with a natural LCS structure with a nontrivial holonomy class.
(2) No. The answer to (1) implies already that $S^1 \times S^3$ admits such a structure, even though it cannot be symplectic since $H^2(S^1 \times S^3) = 0$. In fact, recent work of Bertelson and Meigniez proves an existence h-principle for LCS structures on any almost symplectic manifold in any nontrivial holonomy class (and in fact you can also specify the LCS structure to further live in any specified twisted second cohomology class).
(3) There's a whole story for the ways in which LCS geometry and symplectic geometry are similar and different, and it would take a long time to state all of the interactions. For classical examples and discussion about the group of Hamiltonian diffeomorphisms, see e.g. work of Banyaga and Haller and Rybicki, the latter of which proves that the group of LCS-Hamiltonian diffeomorphisms is simple.
In response to a comment below (this was too long to comment back), my understanding of the historical motivations is as follows (see also this survey of Bazzoni for more):
(1) E. Cartan was interested in classifying pseudo-groups (1909), which might be considered models for geometry. From his classification naturally popped out (locally) conformal symplectic geometry. (Remark: This may be considered pre-history, since I'm not sure that Cartan himself really studied anything about the field itself, and his work isn't really mentioned in the LCS literature at all.)
(2) The term "symplectic" had already been coined in the late 1930's (e.g. Weyl), but the corresponding "flat" geometry that we now call symplectic geometry had not yet been invented until, as far as I'm aware, Hwa-Chung Lee's 1943 paper. Incredibly, Lee defines not just the “flat” geometry, but the “conformally flat” geometry. Hence, it is actually a historical accident that LCS geometry has remained relatively hidden under the shadows of standard symplectic geometry – the notion of an LCS manifold was developed at the same time!
(3) The resurgence of LCS geometry in the mid-70’s was largely due to work of Vaisman in a series of papers starting with one from 1976, and tended to focus quite a bit on extra metric structure (e.g. locally conformal Kaehler (LCK) manifolds). It is worth noting that such metric aspects, which are a little askew from the more differential topological discussion above, still maintains a vibrant research community.
(4) It is worth saying that some of the most basic tools in symplectic geometry have been imported into the LCS world. I mentioned the Haller-Rybicki work on the simplicity of the LCS Hamiltonian group as an example. The Moser trick was imported by Bande and Kotschick, and the Moser trick can be made relative and parametric so as to give corresponding neighborhood theorems, as in work of Lê and Oh (for coisotropics) and Otiman and Stanciu (more generally).
(5) More recent renewed interest (including my own) was largely motivated by work of Eliashberg and Murphy, where they prove a precursor to the aforementioned Bertelson-Meigniez result, which uses many of the same tools. One hope for the future is to be able to import both rigid and flexible techniques from symplectic geometry into LCS geometry.
Best Answer
If $\mathcal{L}_v J= 0$ and $\mathcal{L}_v \Omega= 0$, then $\mathcal{L}_v g=0$ so $v$ is a Killing vector field. Indeed, the property $\mathcal{L}_v J= 0$ is equivalent to the condition that the (local) flow of $v$ preserves $J$ and the property $\mathcal{L}_v \Omega= 0$ is equivalent to the condition that the (local) flow of $v$ preserves $\Omega$. Since $g$ is costructable by $J$ and $\Omega$, if both $J$ and $\Omega$ are preserved by the flow, then $g$ is preserved by the flow as well and the vector field is Killing.
Hamiltonian vector fields have the property $\mathcal{L}_v \Omega= 0$. If your definition of ``real-holomorphic'' implies $\mathcal{L}_v J= 0$, then the answer on your question is yes, yes, yes, ya, ya, ya.