The answer is yes. Let $a$ be a primitive root of $p$. Then $a+kp$ is a primitive root of $p$ for every $k$. By Dirichlet's Theorem on primes in arithmetic progression, there are infinitely many primes of the form $a+kp$.
Remark: The problem of whether for any prime $p\gt 2$ there is a primitive roots $a$ with $2\le a\le p-1$ is open. I believe that the best unconditional result is that for large enough $p$, the least prime primitive root of $p$ is $\lt p^k$, for a constant $k\approx 3$. Under reasonable but unproved hypotheses (versions of GRH) one can bring this down to a power of $\log p$.
One can find some information, and references, in Greg Martin's The Least Prime Primitive Root and the Shifted Sieve.
You may also want to look at this paper, sadly behind a pay wall. There is a fair literature on the subject.
Here is one way to do it that requires knowing only a primitive root $\bmod p$ to get all primitive roots $\bmod p^2$, provided that $p$ is twice a smaller prime plus $1$, so that all nonzero residues besides $\pm 1$ are either quadratic or primitive. I will demonstrate it for $p=5$ letting you apply it for $p=11$.
Start with $2$ as a primitive root $\bmod 5$. Then there will be a series of primitive roots $\equiv 5k+2 \bmod 25$. For one residue $k \bmod 5$ we will have $(5k+2)^4\equiv 1 \bmod 25$ and that value of $k$ must be rejected, but other values of $k$ will all work.
Use the Binomial Theorem to get $(5k+2)^4$, but retain only the first ad zero powers of $k$ (why?). Thus
$(5k+2)^4 \equiv (4)(5k)(2^3)+2^4 \equiv 1 \bmod 25$
$10k+16 \equiv 1, 10k \equiv 10 \bmod 25$
Divide by $10$ noting that the modulus is to be divided by $gcd(10,25)=5$. Then $k\equiv 1 \bmod 5$ meaning $7$ will not be a primitive root $\bmod 25$ but all other residues two greater than a multiple of $5$ will work. We therefore get a subset of primitive roots:
$\{2,12,17,22\}$
Now take the root we rejected, $7$. If we multiply the above primitive roots by it, we get nonprimitive roots because these are quadratic residues. But multiply by $7$ twice, that is by $7^2$, and we get another set of primitive roots because the products are non-quadratic and also miss being $\equiv -1 \bmod 5$. Since $7^2 \equiv 24 \equiv -1 \bmod 25$ this gives another subset of primitive roots
$\{3,8,13,23\}$
Multiplying by $7^2$ again cycles us back to the first subset, so the complete set of primitive roots $\bmod 25$ will be the union of the two subsets
$\{2,3,8,12,13,17,22,23\}$.
Now apply this method to$11=2×5+1$ and see what happens. Since $11$, like $5$, is not one greater or one less than a multiple of $8$, $2$ will be a primitive root $\bmod 11$. Since there are four such primitive roots $\bmod 11$ overall you will need to generate four subsets before taking their union.
Best Answer
3 is a primitive root mod $43^{63}$. To see this, note that $\phi(43^{63}) = 2\cdot 3\cdot 7 \cdot 43^{62}$, and $3^{3\cdot 7 \cdot 43^{62}} \ne 1$, $3^{2\cdot 7\cdot 43^{62}} \ne 1$, $3^{2\cdot 3\cdot 43^{62}} \ne 1$, and $3^{2\cdot 3\cdot 43^{61}} \ne 1$, mod $43^{63}$.
In general, tests for primitive roots are fast to perform once you've factored $\phi(n)$. Since primitive roots are relatively frequent among all residue classes, simply randomly testing numbers (or testing small numbers until one is found), is a viable algorithm to find primitive roots.