The answer is <b>Yes</b>.
Theorem. The following are equivalent for any Hausdorff
space X.
1. X is compact.
2. X<sup>κ</sup> is Lindelöf for any cardinal
κ.
3. X<sup>ω<sub>1</sub></sup> is Lindelöf.
Proof. The forward implications are easy, using Tychonoff
for 1 implies 2, since if X is compact, then
X<sup>κ</sup> is compact and hence Lindelöf.
So suppose that we have a space X that is not compact, but
X<sup>ω<sub>1</sub></sup> is Lindelöf. It
follows that X is Lindelöf. Thus, there is a countable
cover having no finite subcover. From this, we may
construct a strictly increasing sequence of open sets
U<sub>0</sub> subset U<sub>1</sub> subset ... U<sub>n</sub>
... with the union U{ U<sub>n</sub> | n in ω} = X.
For each J subset ω<sub>1</sub> of size n, let U<sub>J</sub> be
the set {s in X<sup>ω<sub>1</sub></sup> | s(α)
in U<sub>n</sub> for each α in J}. As the size of J
increases, the set U<sub>J</sub> allows more freedom on the
coordinates in J, but restricts more coordinates. If J has
size n, let us call U<sub>J</sub> an open n-box, since it
restricts the sequences on n coordinates. Let F be the
family of all such U<sub>J</sub> for all finite J subset
ω<sub>1</sub>.
This F is a cover of X<sup>ω<sub>1</sub></sup>. To
see this, consider any point s in
X<sup>ω<sub>1</sub></sup>. For each α in
ω<sub>1</sub>, there is some n with s(α) in
U<sub>n</sub>. Since ω<sub>1</sub> is uncountable,
there must be some value of n that is repeated unboundedly
often, in particular, some n occurs at least n times. Let J
be the coordinates where this n appears. Thus, s is in
U<sub>J</sub>. So F is a cover.
Since X<sup>ω<sub>1</sub></sup> is Lindelöf,
there must be a countable subcover F<sub>0</sub>. Let J* be
the union of all the finite J that appear in the
U<sub>J</sub> in this subcover. So J* is a countable subset
of ω<sub>1</sub>. Note that J* cannot be finite,
since then the sizes of the J appearing in F<sub>0</sub>
would be bounded and it could not cover
X<sup>ω<sub>1</sub></sup>. We may rearrange indices
and assume without loss of generality that J*=ω is
the first ω many coordinates. So F<sub>0</sub> is
really a cover of X<sup>ω</sup>, by ignoring the
other coordinates.
But this is impossible. Define a sequence s in
X<sup>ω<sub>1</sub></sup> by choosing s(n) to be
outside U<sub>n+1</sub>, and otherwise arbitrary. Note that
s is in U<sub>n</sub> in fewer than n coordinates below
ω, and so s is not in any n-box with J subset
ω, since any such box has n values in U<sub>n</sub>.
Thus, s is not in any set in F<sub>0</sub>, so it is not a
cover. QED
In particular, to answer the question at the end, it suffices to take any uncountable kappa.