50% is wrong: the closer you can get to 100% the better off you are.
Let the measure of the target region $T$ be $t$ and the measure of the enclosing (or "probe") region $V$ be $v$. The chance of a uniformly random point in $V$ to lie in $T$ therefore is $t/v$. This Bernoulli distribution has variance
$$\frac{t}{v}\left(1-\frac{t}{v}\right).$$
The estimate of $t$ is based on $n$ independent uniform samples of $V$, which is therefore a Binomial$(n, t/v)$ variate with variance $n$ times greater than that of a single sample. When its outcome is $X$ the estimate will be the proportion $\hat{t}_n = v\left(\frac{X}{n}\right) = \left(\frac{v}{n}\right)X$. Its variance is
$$\text{Var}(\hat{t}_n) = \left(\frac{v}{n}\right)^2\text{Var}(X) = \left(\frac{v}{n}\right)^2 n\left(\frac{t}{v}\left(1-\frac{t}{v}\right)\right) = \frac{1}{n}t(v-t).$$
Because $V$ encloses $T$, $v\ge t$. The variance, being a linear function of $v$ in this interval, obviously is minimized at $v=t$. Ergo, the correct rule is to find an enclosing volume that is as close to $T$ as possible. Ideally, $V$ will be $T$ itself and the correct answer will be available upon taking $n=0$ samples!
As a check, I performed $500$ Monte-Carlo integrations of $\int_2^3\int_6^7 x y\ dy dx$ using enclosing volumes of constant heights $21$, $32.5$, and $84$ and $n=1000$ iterations per integration. For instance, here are the results of one of the integrations for $21$ (showing the target region $T$ beneath the blue surface graphing $xy$ over the square $[2,3]\times[6,7]$). The black dots fall within $T$ while the red dots, although still within $V$, fall outside $T$.

The results of these $500$ trials are
Height: 21 32.5 84
------------------------------
t/v: 77% 50% 19%
Mean: 16.258 16.235 16.280
Variance: 0.077 0.271 1.061
t(v-t)/n: 0.077 0.264 1.100
The true mean of $65/4$ was adequately estimated on average in all three situations. Their variances are comfortably close to the value of $t(v-t)/n$ previously derived.
Clearly 50% (the middle column) is not more accurate: its estimated variance of $0.271$ is almost four times worse than the estimated variance of $0.077$ achieved when the target region is $77\%$ of the probe region (left column). It would therefore take approximately $0.271/0.077 = 3.5$ times as many iterations in the $50\%$ configuration to achieve the same level of accuracy as the $77\%$ configuration. In fact, when I redid the $50\%$ calculation using $n=3500$ iterations per integral, the variance of $500$ trials was $0.065$. This does not differ significantly from $0.077$.
The import of this variance calculation is plain: lower variances mean either less computation or better accuracy (or both).