They are not perfectly positively correlated: Even when random variables are deterministically related (which would require $X$ to be deterministic in this case), perfect correlation requires them to be related via an affine transformation. This would require a relationship of the form:

$$V(t) = \ln \Big[ 1+\frac{U(t)}{X(t)} \Big] = a + b U(t),$$

where $a \in \mathbb{R}$ and $b>0$. Solving for the process $X$ gives:

$$X(t) = \frac{U(t)}{\exp(a + b U(t))-1}.$$

This is inconsistent with your specification that $X$ is a geometric Brownian motion.

They are not perfectly positively correlated: Even when random variables are deterministically related (which would require $X$ to be deterministic in this case), perfect correlation requires them to be related via an affine transformation. This would require a relationship of the form:

$$V(t) = \ln \Big[ 1+\frac{U(t)}{X(t)} \Big] = a + b U(t),$$

where $a \in \mathbb{R}$ and $b>0$. Solving for the process $X$ gives:

$$X(t) = \frac{U(t)}{\exp(a + b U(t))-1}.$$

This is inconsistent with your specification that $X$ is a geometric Brownian motion. However, note that if your geometric Brownian motion process has a large mean and small variance (such that it is approximately constant at a mean value $\mu_X$ that is much bigger than $U(t)$) then you would have $X(t) \approx \mu_X \gg U(t)$ which gives the approximation:

$$V(t) = \ln \Big[ 1+\frac{U(t)}{X(t)} \Big] \approx \frac{U(t)}{X(t)} \approx \frac{1}{\mu_X} \cdot U(t),$$

so in this case you could get something that is close to an affine transform, and so you would get something close to perfect correlation.