You are correct in observing that the effective potential is complex for certain values of the background field. This is somewhat of a thorny issue. In principle, when deriving the effective potential one writes it as a Legendre transformation of the generating functional of connected diagrams $ W[J] $. This assumes that the effective potential is convex in the variable $\phi$. When there is spontaneous symmetry breaking present at the classical level this is obviously not true, as evidenced by the negative curvature at the origin of field space. The conventional wisdom is that our whole derivation does not breakdown into nonsense. Instead, when a field dependent mass turns negative then the corresponding field value does not represent a stable state. This is signaled through the effective potential by it acquiring a non-zero imaginary part that plays the role of a decay rate of this state. However, see [Precision decay rate calculations in quantum field theory] for how to calculate the actual physical decay rate properly.
Your second observation, that the logarithm diverges as $ \phi \rightarrow v $ is correct but also nothing to worry about. This is because the prefactor of the logarithm, $ \left(m_G^2\right)^2 $, also goes to zero in this limit (faster than the logarithm diverges).
On a more general note, there is nothing inconsistent about evaluating the effective potential for general field values. However, the effective potential does not in general represent a physical quantity. Only when evaluated at extrema.
I recommend [Consistent Use of Effective Potentials] for a deep treatment of the effective potential and how to derive physical quantities from it.
Any way I look at it yes, apparently the dimreg continuation of that integral only has poles at even dimensions. I'm not sure if the kinetic part of the effective action has divergences or not but that would be a little tiresome to check. If you want to do it anyway, the object to look at is
$$\text{tr}\log(1+(-\partial^2+M^2)^{-1}(V''(\phi)-M^2))=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int\prod_{\alpha=1}^n\frac{d^dk_\alpha d^dx_\alpha}{(2\pi)^d}\frac{V''(\phi(x_\alpha))-M^2}{k_\alpha^2+M^2}e^{ik_\alpha(x_{\alpha-1}-x_\alpha)},$$
with $x_0=x_n$ for some arbitrary and convenient $M^2$, up to second order partial derivatives of $\phi$. This object is what remains of $\text{tr}\log(-\partial^2+V''(\phi))$ after you take out a $\phi$ independent $\text{tr}\log(-\partial^2+M^2)$. If the kinetic term diverges then due to the renormalization of the field itself you're going to get a flow.
Anyway, the point I want to make is about the culmination of your question: how can non-divergent loop corrections give rise to RG flow? Well, remember that RG depends on not only your theory, but the renormalization scheme. If you're working with MS or $\bar{\text{MS}}$ there's surely no flow without divergence, but if you're doing on-shell or some other thing, there probably is.
Why? Because renormalization is not simply a recipe to do away with divergences. The question it aims to solve is basically "if I systematically ignore some degrees of freedom, is there a change of variables I can do to the remaining ones to essentially get back to the same theory? If so, how exactly does it differ from what I started with?"
It's not intrinsically tied to divergences. The way I particularly like to illustrate this is the following. Take a theory of a field $\phi$ with action $S[\phi;\lambda]$ for some parameters $\lambda$. Separate $\phi=(\phi_\text{care},\phi_\text{don't})$ as independent degrees of freedom. The path integral is then
$$\int D\phi\,e^{-S[\phi]}=\int D\phi_\text{care}D\phi_\text{don't}\,e^{-S[\phi_\text{care},\phi_\text{don't};\lambda]}=\int D\phi_\text{care}e^{-S_\text{eff}[\phi_\text{care};\lambda]},$$
where
$$S_\text{eff}[\phi_\text{care};\lambda]=-\log\int D\phi_\text{don't}\,e^{-S[\phi_\text{care},\phi_\text{don't};\lambda]}.$$
The magic of renormalization then happens if there is a change of variables $\phi_\text{care}\mapsto\tilde\phi$ such that
$$S_\text{eff}[\phi_\text{care};\lambda]=S[\tilde\phi;\tilde\lambda]$$
for some new set of parameters $\tilde\lambda$. You have just mapped your coarse grained theory to your original one, and the RG flow is the map $\lambda\mapsto\tilde\lambda$. This is completely independent of the fact that the parameters involved may or may not be infinite.
Translating to traditional particle physics QFT, the step $\phi=(\phi_\text{care},\phi_\text{don't})$ corresponds to choosing a particular regularization (Wilson cutoffs, dimreg), while $\phi_\text{care}\mapsto\tilde\phi$ corresponds to your renormalization conditions (MS, on shell). Try doing renormalization on you divergence free theory with on-shell or zero-energy conditions, see if you get any flow.
Best Answer
The problem here is a problem about renormalization.
First, if we have a Lagrangian $\mathcal{L}$, with only so-called bare parameters, such as $\mathcal{L} = \frac{1}{2}(\partial\phi)^{2}-\frac{1}{2}m^{2}_{0}\phi^{2}-\frac{\lambda_{0}}{4!}\phi^{4}$. We can use this model to calculate observables and compare them with the experiments to get the parameters $(m_{0},\lambda_{0})$ in my model if this model is sufficient. This is the basic logic, parameters in the Lagrangian depend on experiments.
Second, to isolate the UV part physics (or you can say to perturb using the physical parameters), we isolate the Lagrangian as $\mathcal{L} = \mathcal{L}_{1}+\mathcal{L}_{ct}$ where $\mathcal{L}_{ct}$ contains so-called conterterms. As you know, for renormalizable Lagrangian, terms in $\mathcal{L}_{ct}$ are got by isolating parameters in the bare Lagranian, such as $m_{0} = m + \delta_{m}$. If we focus on no-counterterms part $\mathcal{L}_{1}$ only, the parameters in it depend on both experiments and counterterms, equivalently, both experiments and renormalization schemes.
Then, back to your problem. $\phi_{cl}$ can be view as an observable (not really in fact), and it depends on experiments. If we are using the full Lagrangian $\mathcal{L}$, not isolating counterterms, using bare parameters ($(\lambda_{0},\mu_{0})$ here), $\phi_{cl}$ has a specific dependence on (bare) parameters. We can use this dependence and the experiments to get an equation of (bare) parameters. However, if we isolate the counterterms and focus on the $\phi_{cl}$ dependence on the parameters in $\mathcal{L}_{1}$ only ($(\lambda,\mu)$ here). We can have any $\phi_{cl}$ dependence by having different counterterm parameters, equivalently by using different renormalization schemes. The logic is we don't care if $\phi_{cl} = \mu/\sqrt\lambda$, $\phi_{cl}$ depends on experiments, and we use the relation between it and parameters to determine the parameters in Lagrangian.
The first renormalization scheme you present is similar to the on-shell scheme. Namely, the parameters $(\mu,\lambda)$ in $\mathcal{L}_{1}$ (not bare parameters in full $\mathcal{L}$) can be view as "physical parameters". The relation between $\phi_{cl}$ and $(\mu,\lambda)$ is the same as that in classical theory.
The $\overline{MS}$ scheme have a renormalization scale $M$ (use $mu$ in most contexts, but $mu$ has been a parameter in Lagrangian here), the existence of $M$ is part of this renormalization scheme. Following the logic above, if we use $\overline{MS}$ scheme and derive a relation between $\phi_{cl}$ and $(\mu,\lambda,M)$ and the experiment value of $\phi_{cl}$, we can assign a value of $M$ as we like, and get a relation between $\mu$ and $\lambda$ which can help us determine the value of them.