Yes, it is possible. The $R^2$ and the $t$ statistic (used to compute the p-value) are related exactly by:
$
|t| = \sqrt{\frac{R^2}{(1- R^2)}(n -2)}
$
Therefore, you can have a high $R^2$ with a high p-value (a low $|t|$) if you have a small sample.
For instance, take $n = 3$. For this sample size to give you a (two-sided) p-value less then 10% you would need an $R^2$ greater than 85% -- anything less than that would give you "non-significant" p-value.
As a concrete example, the simulation below produces an $R^2$ close to 0.5 with a p-value of $0.516$.
set.seed(10)
n <- 3
x <- rnorm(n, 0, 1)
y <- 1 + x + rnorm(n, 0, 1)
summary(m1 <- lm(y ~ x))
Call:
lm(formula = y ~ x)
Residuals:
1 2 3
-0.36552 0.42802 -0.06251
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.7756 0.4261 1.82 0.320
x 0.5065 0.5333 0.95 0.516
Residual standard error: 0.5663 on 1 degrees of freedom
Multiple R-squared: 0.4743, Adjusted R-squared: -0.05148
F-statistic: 0.9021 on 1 and 1 DF, p-value: 0.5164
For the opposite case (low p-value with low $R^2$), you can trivially obtain that by setting a regression where $x$ has a low explanatory power and let $n \to \infty$ to get a p-value as small as you want.