Here is a cheaper alternative depending on what you mean by a modification. Consider $L(z)$ any Dirichlet L function different from $\zeta$.
Joint universality theorem: Let $K$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$ with connected complement. For any two functions $f_1$ and $f_2$ holomorphic in the interior of $K$ (vanishing or not) and every $\epsilon>0$, we have that the limit $$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \{ t \leq T: \sup |f_1(z) - \log \zeta(z +i t)| + \sup |f_2(z) - \log L(z +i t)| < \epsilon\} $$ is positive for $\lambda$ being the Lebesgue measure.
From this, we can deduce:
Corollary: Let $K_0$ be a compact set in the right half of the critical stripe $1/2< \Re s<1$. Let $f$ be a continuous function on $K_0$, which is holomorphic on an open set containing $K_0$. For every $\epsilon_0>0$, we have that the limit $$ \inf\lim\limits_{T \rightarrow \infty} \frac{1}{T} \lambda \Big\{ t \leq T: \sup\limits_{z \in K_0} \left| f(z) - \frac{\log \zeta(z +i t)}{\log L(z+ it)}\right| < \epsilon_0\Big\} $$ is positive for $\lambda$ being the Lebesgue measure.
Proof: By Runge's theorem, it is sufficient to approximate rational functions, whose poles lie outside of $K_0$. Let $p(z)$ and $q(z)$ be polynomials such that $q$ does not vanish on $K_0$. Consider $\epsilon_0>0$ sufficiently small (to be made precise as we go on).
Let $K :=\mathbb{C}-O$, where $O$ is the unbounded, connected component of $\mathbb{C}-K_0$. Consider $\epsilon>0$ sufficiently small, then use the joint universality theorem for $f_1(z)=p(z)$ and $f_2(z) =q(z)$.
We want to show that $$\sup | f_1/f_2(z) - \frac{\log \zeta}{\log L}(z+i t) |< \epsilon_0.$$
We estimate the left-hand side: $$ \leq \sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | + \sup | \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)|.$$
The first summand is easy to estimate: $$\sup | f_1/f_2(z) - \frac{\log \zeta(z+it)}{f_2(z)} | \leq \sup_{z \in K_0} \left| f_2(z)^{-1} \right| \epsilon.$$ The second one is a little bit harder: $$ \sup \Big| \frac{\log \zeta(z+it)}{f_2(z)} - \frac{\log \zeta}{\log L}(z+i t)\Big| \leq $$ $$ \sup \Big| \frac{\log \zeta(z+i t)}{f_2(z)\log L(z+i t)} \Big| \sup | \log L(z+i t) -f_2(z) | < \sup \Big| \frac{\log \zeta(z+i t)}{f_2(z) \log L(z+i t)} \Big| \epsilon,$$ because we have to estimate $$ \sup | \frac{\log \zeta}{\log L}(z+i t) | $$ uniformly in $t$.
This is indeed possible, we have that $$\sup | f_2(z) | - \sup | \log L(z + i t) | < \epsilon$$ and $$ \sup | \log L(z \zeta(z + i t) | - \sup | f_1(z) | < \epsilon$$ by the reversed triangle inequality. So for $\epsilon \leq \sup | f_2(z) |/2$ and $\epsilon \leq \sup | f_1(z) |$ , we have that $$ \sup | \log L(z + i t) | > \sup | f_2(z) |/2$$ and $$ \sup | \log L(z \zeta(z + i t) | < 2 \sup | f_1(z) | .$$ So $$\epsilon_0 := \max\{ \frac{1}{2} \sup |f_2^{-1}| \epsilon, \frac{1}{2} 4* \sup |f_1f_2^{-2}| \epsilon \}$$ will do.
This finishes the proof of the corollary assuming the Joint universality theorem.
