For the Euclidean path integral to be convergent, the Boltzmann factor should be an exponentially decaying function of the scalar field $\phi$. This in turn dictates that the direction of the Wick rotation. (It should perhaps be stressed that Wick rotation is not merely a renaming of time variables, but that it involves an actual deformation of the time-integration contour in the complex time plane.) Standard conventions for the Wick rotation are
$$\begin{align}
-S_E~=~&iS_M, \cr
t_E~=~&it_M, \cr
{\cal L}_E~=~&-{\cal L}_M,\end{align} \tag{1} $$
where the subscripts $E$ and $M$ stand for Euclid and Minkowski, respectively. In OP's case
$$\begin{align} {\cal L}_M~=~&i\phi^{\ast} \frac{d\phi}{dt_M}- {\cal V} ,\cr
{\cal L}_E~=~&\phi^{\ast} \frac{d\phi}{dt_E}+ {\cal V} ,\cr
{\cal V}~=~&\frac{1}{2m}\nabla\phi^{\ast}\cdot \nabla\phi .\end{align} \tag{2}$$
See also this related Phys.SE post.
Proposition. Let there be given two complex numbers $a,b\in \mathbb{C}$ such that ${\rm Re}(a)\geq 0$. In the case ${\rm Re}(a)=0$, we demand furthermore that ${\rm Im}(a)\neq 0$ and ${\rm Re}(b)=0$. The Gaussian integral is well-defined and is given by
$$ \underbrace{\int_{\mathbb{R}}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{\mathbb{R}}(a,b)}
~=~\lim_{\begin{array}{c} x_i\to -\infty \cr x_f\to \infty \end{array} } \underbrace{\int_{[x_i,x_f]}\!dx~ e^{-\frac{a}{2}x^2+bx}}_{=:~ I_{[x_i,x_f]}(a,b)}
~=~\underbrace{\sqrt{\frac{2\pi}{a}}e^{\frac{b^2}{2a}}}_{=:~ F(a,b)}, \tag{A}$$
where it is implicitly understood that the square root has positive real part.
Remark: The Riemann/Darboux integral is not defined for non-bounded sets, so it can only be used for the middle expression of eq. (A).
I) Sketched proof in case of ${\rm Re}(a)> 0$: The function $g(x)=e^{-\frac{{\rm Re}(a)}{2}x^2+{\rm Re}(b)x}$ serves as a majorant function for Lebesgue's dominated convergence theorem, which establishes the first equality of eq. (A). For the second equality of eq. (A), we divide the proof into cases:
Case $a>0$ and $b\in \mathbb{R}$. Complete the square. $\Box$
Case $a>0$. Complete the square. Shift the integration contour appropriately to a horizontal line in the complex plane in order to reduce to case 1, cf. Cauchy's integral theorem. Argue that contributions at infinity vanish. $\Box$
Case ${\rm Re}(a)> 0$. Rotate the integration contour to a line of steepest descent in order to reduce to case 2, cf. Cauchy's integral theorem. Argue that contributions at infinity vanish. $\Box$
II) Sketched proof in the oscillatory case ${\rm Re}(a)=0, {\rm Im}(a)\neq 0, {\rm Re}(b)=0$: The lhs. of eq. (A) is not Lebesgue integrable. It is an improper integral defined via the middle expression of eq. (A). It remains to prove the second equality of eq. (A). It is possible to give a proof using Cauchy's integral theorem along the lines of Jack's answer. In this answer we will instead give a proof in the spirit of an infinitesimal deformation prescription.
Given $\varepsilon>0$. As $x_i\to \infty$ and $x_f\to \infty$ it is not hard to see that $I_{[x_i,x_f]}(a,b)$ oscillates with smaller and smaller amplitude that tends to zero, and it is hence convergent without any regularization. The convergence improves if we let $a$ have a positive real part. In other words, the convergence is uniform wrt. ${\rm Re}(a)\geq 0$, i.e.
$$ \exists X_i,X_f\in \mathbb{R} ~\forall x_i\leq X_i~\forall x_f\geq X_f ~\forall {\rm Re}(a)\geq 0:~~
\left| I_{[x_i,x_f]}(a,b)- I_{\mathbb{R}}(a,b)\right|
~\leq~\frac{\varepsilon}{4}.\tag{B}$$
Next use Lebesgue's dominated convergence theorem with majorant function of the form $g(x)=C~1_{[x_i,x_f]}(x)$ (where $C>0$ is an appropriate constant) to argue that
$$I_{[x_i,x_f]}( i{\rm Im}(a),b)
~=~\lim_{{\rm Re}(a)\to 0^+} I_{[x_i,x_f]}(a,b) , \tag{C}$$
i.e. $\exists {\rm Re}(a)>0$ such that
$$ \left| I_{[x_i,x_f]}( i{\rm Im}(a),b)-I_{[x_i,x_f]}( a ,b) \right|
~\leq~\frac{\varepsilon}{4},\tag{D}$$
and
$$\left| \underbrace{F( a ,b)}_{=~ I_{\mathbb{R}}(a,b)}- F( i{\rm Im}(a),b) \right| ~\leq~\frac{\varepsilon}{4}.\tag{E}$$
In eq. (E) we used that the function $F$ is continuous.
All together, eqs. (B), (D) & (E) yield
$$ \begin{align}
\left| I_{\mathbb{R}}(i{\rm Im}(a),b) - F( i{\rm Im}(a),b)\right|
~\leq~&\left| I_{\mathbb{R}}(i{\rm Im}(a),b)- I_{[x_i,x_f]}( i{\rm Im}(a),b)\right|\cr
&+\left| I_{[x_i,x_f]}( i{\rm Im}(a),b) - I_{[x_i,x_f]}( a ,b)\right| \cr
&+\left| I_{[x_i,x_f]}(a,b)- I_{\mathbb{R}}(a,b)\right|\cr
&+\left|F( a ,b) - F( i{\rm Im}(a),b)\right| \cr
~\leq~&\varepsilon.\end{align} \tag{F}$$
Eq. (F) shows that the second equality of eq. (A) holds. $\Box$
Best Answer
As noted in the edit, I also believe the two results will be the same. Let's do an example using a convergent integral, taking $I\equiv \int d^4k \, f(k^2),$ where $f(k^2)=(k^2-\Delta+i\epsilon)^{-3}$. Then performing the Wick rotation yields $I = -2\pi^2i\int_0^\infty \frac{k^3dk}{(k^2+\Delta)^3}=-\frac{i}{2}\frac{\pi^2}{\Delta}.$
On the other hand, direct integration over $k^0$ yields $I = -\frac{3\pi i}{8}4\pi\int_0^\infty \frac{k^2dk}{(k^2+\Delta)^{5/2}} = -\frac{i}{2}\frac{\pi^2}{\Delta}.$