Let $U:=\{z\in\mathbb{C}\ |\ \operatorname{Im}(z)>0\}$. Let $h(U,\mathbb{R})$ be the set of real harmonic functions defined on $U$. Define $$h_0(U,\mathbb{R}):=\{u\in h(U,\mathbb{R})\ |\ \lim_{y\rightarrow+\infty}\sup_{x\in\mathbb{R}}|u(x+iy)| =0 \}.$$
Is it true that for every $u\in h_0(U,\mathbb{R})$, if $v\in h(U)$ is a conjugate harmonic of $u$, then there exists $c\in\mathbb{R}$ such that $v+c\in h_0(U,\mathbb{R})$ ?
The reason that made me wonder about this question is that if $f\in L^p(\mathbb{R})$ with $1\le p<+\infty$ and $u(x+iy):=(P_y*f)(x)$, where $(P_y)_{y>0}$ is the Poisson kernel for the upper half plane, then $u\in h_0(U,\mathbb{R})$ and if $v(x+iy):=(Q_y*f)(x)$, where $(Q_y)_{y>0}$ is the conjugate Poisson kernel for the upper half plane, then $v$ is a conjugate harmonic for $u$ and $v\in h_0(U,\mathbb{R})$.
However, for the general case of $u\in h_0(U,\mathbb{R})$ I've no clue about how to attack the problem.
Best Answer
Another approach:
Claim: There exists $f=u+iv$ holomorphic in $U$ such that as $z\to 0$ in $U,$ $u(z) \to -\infty,$ $v(z)\to 0.$
Suppose the claim is proved and we have such an $f.$ Then the function $-if(-1/z) = v(-1/z) -iu(-1/z)$ is holomorphic in $U.$ As $z\to \infty$ within $U,$ $-1/z\to 0$ within $U.$ Hence $v(-1/z) \to 0$ and $-u(-1/z)\to \infty.$ Thus we have a counterexample.
Proof of claim: Consider the functions
$$f_n(z) =\log (z+i/e^n) = \log |z+i/e^n| + i\text {arg }(z+i/e^n),\,\,n=1,2,\dots,$$
where $\log $ denotes the principal value logarithm. These functions are holomorphic in $U$ and are uniformly bounded on compact subsets of $U.$ For $z\in U,$ define
$$f(z)=\sum_{n=1}^{\infty}\frac{f_n(z)}{n^2}.$$
By Weierstrass M, the series converges uniformly on compact subsets of $U,$ hence $f$ is holomorphic in $U.$ Writing $f=u+iv,$ we have
$$u(z) = \sum_{n=1}^{\infty} \frac{\log |z+i/e^n|}{n^2}.$$
Note all summands are negative on $\{z\in U: |z|<1/2\}.$ Thus for any $N,$
$$\limsup_{z\to 0} u(z) \le \lim_{z\to 0} \sum_{n=1}^{N} \frac{\log |z+i/e^n|}{n^2} =\sum_{n=1}^{N} \frac{\log |i/e^n|}{n^2} = - \sum_{n=1}^{N} \frac{1}{n}.$$
Since $N$ is arbitrary, we see $\lim_{z\to 0} u(z)=-\infty.$ (Just to be clear, these limits are taken as $z\to 0$ within $U.$)
As for $v(z),$ we have
$$v(z) = \sum_{n=1}^{\infty} \frac{\text {arg }(z+i/e^n)}{n^2}.$$
This series actually converges uniformly on all of $U.$ Thus
$$\lim_{z\to 0} v(z) = \sum_{n=1}^{\infty} \lim_{z\to 0}\frac{\text {arg }(z+i/e^n)}{n^2} = \sum_{n=1}^{\infty} \frac{\text {arg }(i/e^n)}{n^2} = \sum_{n=1}^{\infty} \frac{\pi/2}{n^2}.$$
We're not quite done: Letting $c$ denote the last sum, we see $f(z)-ic$ has the claimed properties.
Previous answer: I'm pretty sure this is false. Define $V=\{x+iy: x>0, |y|<x^2.$ Then $V$ is simply connected, so there is a conformal map $g:U\to V.$ And we should be able to arrange things so that $z\to \infty$ in $U$ iff $g(z)\to 0$ in $V.$ Now define
$$h(z) = -i\log g(z)= \text {arg }g(z) - i\ln |g(z)|.$$
As $z\to \infty$ in $U,$ $\text {arg }g(z) \to 0.$ That's because $g(z)\to 0$ tangent to the real axis. But the conjugate function $-\ln |g(z)| \to \infty.$