As a strengthening of what @KConrad commented, it would imply that the density of nontrivial zeros on the critical line is 100% in each horizontal strip of height 1, which is not useless: this is equivalent to the Lindelöf Hypothesis, which states that $\zeta \left( \frac{1}{2} + i t \right) = \mathcal{O}_{\varepsilon} \left( 1 + \lvert t \rvert^{\varepsilon} \right)$.
One example of a consequence of the Lindelöf Hypothesis (which is exactly much easier to prove directly from your non-effective Riemann Hypothesis using the explicit formula) is that the prime gaps satisfy $p_{n + 1} - p_n \leq \sqrt{p_n} \log \left( p_n \right)^2$, improving the best current unconditional result by Baker-Harman-Pintz of $p_{n}^{0.525}$.
The exponent of the logarithm might be a bit less, but I decided to err on the side of caution. By the way, this is still very far from the conjectured upper bound $p_{n}^{\varepsilon}$ (and in fact it is relatively widely believed that the gap is at most $C \log \left( p_n \right)^2$ for some absolute constant $C$).
However, the Lindelöf Hypothesis appears in estimating arithmetic sums, as many counting problems can be transformed into a zeta integral. Let me illustrate by a simple example. We will prove (conditionally) the following:
$$\sum_{n = 1}^{N} d (n) = n \log n + (2 \gamma - 1) n + \mathcal{O}_{\varepsilon} \left( n^{1/2 + \varepsilon} \right)$$
where $d(n)$ is the number of divisors of $n$ and $\gamma$ is the Euler-Mascheroni constant. This is usually proved via the Dirichlet hyperbola method (and with good reason), and we get a slightly weaker error term than usual, but this is just to illustrate the technique. Recall the classical inverse Mellin transform
$$\intop_{c - i \infty}^{c + i \infty} x^s \frac{\mathrm{d} s}{s} = 1_{x > 1} + \frac{1}{2} 1_{x = 1}$$
for any $c > 0, \ x \in \mathbb{R}$. Taking $c > 1$, and interchanging summation and integration we get (up to an error of $\frac{d (n)}{2}$, which is negligible)
$$\sum_{n = 1}^{N} d(n) = \intop_{c - i \infty}^{c + i \infty} \sum_{n = 1}^{\infty} d(n) \left( \frac{N}{n} \right)^{s} \frac{\mathrm{d} s}{s} = \intop_{c - i \infty}^{c + i \infty} \zeta \left( s \right)^2 \frac{N^s \mathrm{d} s}{s}$$
Now, shift the contour to $\mathrm{Re(s) = \frac{1}{2}}$. The residue picked up at the pole $s = 1$ is exactly $N \log N + (2 \gamma - 1) N$, so all we have left to proveis to show that the expression
$$N^{\frac{1}{2}} \intop_{-\infty}^{\infty} \zeta \left( \frac{1}{2} + i t \right)^2 N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$
is $\mathcal{O}_{\varepsilon} \left( N^{1/2 + \varepsilon} \right)$, or equivalently that the integral is $\mathcal{O}_{\varepsilon} \left( N^{\varepsilon} \right)$.
Here is the point where I tell you that I actually lied beforehand: it turns out that using the full inverse Mellin transform is, although very elegant, not necessarily the best choice to get a good analytic bound. What is usually done is approximate it by integrating not from $c - i \infty$ to $c + i \infty$, but from $c - i N$ to $c + i N$, where $c$ is say something like $1 + \frac{1}{\log N}$. I don't remember the details off the top of my head (they appear for example in Montgomery's book, and in a few expositions of proofs of the Prime Number Theorem), so just trust me here when I say that it is sufficient to bound the integral
$$\intop_{- N}^{N} \zeta \left( \frac{1}{2} + i t \right)^{2} N^{i t} \frac{\mathrm{d} t}{\frac{1}{2} + i t}$$
But now (and here we finally use the Lindelöf Hypothesis!) we can bound pointwise this integral, and get that it is $\mathcal{O}_{\varepsilon} \left( N^{\varepsilon} \right)$ as required.
This example, although somewhat stupid, shows the power of the Lindelöf Hypothesis. Indeed, see Tao's answer The relationship between the Dirichlet Hyperbola Method, the prime counting function, and Mertens function, where he points out the fundamental difference between arithmetic functions with zeta in their denominator (whose behaviour is controlled very much by the zeroes of zeta) and arithmetic functions with zeta in the numerator. Despite that, we still managed to use information about the zeroes of zeta to get a nontrivial estimate.
Best Answer
As long as a result remains unproved it can be pure speculation about whether it is really hard or just nobody has found the right simple idea, though in this case it seems plausible that the desired nonvanishing is going to be require a profound new idea. (For more on the task of deciding if a problem is hard before it has been solved, look at the answers to the question https://math.stackexchange.com/questions/88709/complex-math-problem-that-is-easy-to-solve.) That the substantial numerical evidence behind RH should somehow be related to mathematicians being able to establish a uniform zero-free strip $1-\varepsilon < {\rm Re}(s) \leq 1$ is sort of like living in a dream world: mere numerical evidence in this case doesn't suggest any idea of how to create a zero-free vertical strip. It doesn't even suggest how to create a zero-free vertical line: look at any proof that $\zeta(s) \not= 0$ on the line ${\rm Re}(s) = 1$ and tell us how numerical data would lead you to such a proof!
If I had to give one simple reason that it's not (apparently) easy to construct a zero-free vertical strip, I'd say it's because the line ${\rm Re}(s) = 1$ is not compact. That is, although we know $\zeta(s) \not= 0$ on ${\rm Re}(s) = 1$, it doesn't immediately follow that $\zeta(s) \not= 0$ on a uniform strip $1-\varepsilon < {\rm Re}(s) \leq 1$ because the line ${\rm Re}(s) = 1$ is not compact.
To see why lack of compactness is relevant, consider the analogues of $\zeta(s)$ defined over finite fields. (Their values are complex, but the data used to construct them relies on geometry in characteristic $p$.) They are power series in $p^{-s}$ where $p$ is the characteristic of the finite field and ${\rm Re}(s) > 1$, and they are in fact rational functions in $p^{-s}$, which provides a meromorphic continuation to $\mathbf C$. Being series in $p^{-s}$, these zeta-functions are periodic in $s$ with period $2\pi{i}/\log p$ and therefore they can be regarded equivalently as power series in the variable $z = p^{-s}$. In the $z$-world, the line ${\rm Re}(s) = 1$ corresponds to the circle $|z| = 1/p$ and the right half-plane ${\rm Re}(s) \geq 1$ corresponds to the disc $|z| \leq 1/p$. (It really corresponds to the punctured disc $0 < |z| \leq 1/p$, but we can fill in the value of the function at the origin using its constant term.) Any circle $|z| = r$ or disc $|z| \leq r$ is compact, so a continuous function on ${\mathbf C}$ that is nonvanishing on $|z| \leq r$ is necessarily nonvanishing on $|z| \leq r+\varepsilon$ for some $\varepsilon > 0$. That extra uniformly larger disc translates back in the $s$-world into an added uniform strip where the function is nonvanishing.
Returning to $\zeta(s)$, since it is nonvanishing on the line ${\rm Re}(s) = 1$ there must be some open set containing that line where $\zeta(s)$ is nonvanishing, but there's no reason you can deduce without some new idea that this open set containing ${\rm Re}(s) = 1$ must contain a strip $1 - \varepsilon < {\rm Re}(s) \leq 1$ for some $\varepsilon > 0$. If ${\rm Re}(s) = 1$ were compact then that deduction would be easy: all open sets containing that line would contain such a strip. (You could make the same argument using the closed half-plane ${\rm Re}(s) \geq 1$ in place of the line ${\rm Re}(s) = 1$: an open set containing ${\rm Re}(s) \geq 1$ doesn't have to contain a half-plane ${\rm Re}(s) > 1-\varepsilon$ for some $\varepsilon > 0$, although it would if ${\rm Re}(s) \geq 1$ were compact.)
Here's another reason that the existence of zero-free vertical strips will probably be hard to prove: there are Dirichlet series that look similar to the $\zeta$-function (e.g., they converge on ${\rm Re}(s) > 1$ and have an analytic continuation and functional equation of a similar type) but they have lots of zeros to the right of the critical line, and can even have zeros in the half-plane ${\rm Re}(s) > 1$. These examples are called Epstein zeta-functions, and unlike $\zeta(s)$ they do not have any known Euler product. Somehow the Euler product for $\zeta(s)$ is probably going to play a crucial role in making progress on an understanding of its zero-free vertical regions.